Thinking
Obviously to DP. Set \ (dp_ {u, i} \) represents \ (U \) within a subtree containing \ (U \) communication block XOR out \ (I \) is the number of programs found that transfer can FWT optimization written as generating function is this:
\ [dp_ {U} = X ^ {val_u} \ Prod (dp_v +. 1) \]
Finally, the answer is that all DP values and, thereby obtaining a simple \ (O (nmQ) \) of practice. (All intermediate computation value represented by dots)
Obviously use a dynamic optimization of the DP, we note a further \ (S_u \) represents the value of DP and its subtree and DP values, written in the form of a matrix, is
\ [\ left [\ begin { matrix} dp_u \ \ S_u \\ 1 \ end {matrix } \ right] = \ left [\ begin {matrix} dp'_u & 0 & dp'_u \\ dp'_u & 1 & S'_u \\ 0 & 0 & 1 \ end {matrix} \ right] \ times \ left [ \ begin {matrix} dp_v \\ s_v
\\ 1 \ end {matrix} \ right] \] significance (metastasis: \ (dp_u + = dp'_u dp'_udp_v, S_u = S'_u-dp'_u + dp_u s_v is + \) )
Of course, this can only be used for a son, the son of a plurality of time still can not take up directly to the matrix.
Consider sectional chain, the information is combined with the presence of light son matrix, the matrix on the heavy chain to accelerate the transfer, on the right.
How to modify it?
For the point to be modified, see transfer equation above, we found that only \ (x ^ {val_u} \ ) has changed, and the new changed before it is removed.
For the above point, is to be changed back \ (dp_v + 1 \) , it is also to get rid of the original with a new one.
In addition to time may be other than 0, so the number to be replaced \ (x \ times mod ^ y \) recording format to do division. Once this thing two numbers together on the bombing, but you find only \ (dp_u \) need to do division, and multiplication and division, and only about, so there is no problem.
But you do constant and explode, so you need to find matrix \ (\ left [\ begin { matrix} a & 0 & b \\ c & 1 & d \\ 0 & 0 & 1 \ end {matrix} \ right] \) multiplication have closure, so long as the maintenance four values, fast.
DP start when there is a detail. When the transfer of the multiplication of brackets must not open, or you gone.
When the output of a detail answer. Normal transition matrix should be back to take a vector to get the correct solution, but you find the right one transition matrix is the vector, so you can directly use all the matrices are multiplied together after one of the right information. Note that the left this time \ (a, c \) two elements already do not know what things. (Beginner dynamic DP when Mongolia in it for a long time)
(This question when is a dynamic review DP, after all, and not so much ...... FWT)
Code
Since my space explode, we need shortkeep things ......
#include<bits/stdc++.h>
clock_t t=clock();
namespace my_std{
using namespace std;
#define pii pair<int,int>
#define fir first
#define sec second
#define MP make_pair
#define rep(i,x,y) for (int i=(x);i<=(y);i++)
#define drep(i,x,y) for (int i=(x);i>=(y);i--)
#define go(x) for (int i=head[x];i;i=edge[i].nxt)
#define templ template<typename T>
#define sz 30303
#define SS 130
#define mod 10007
typedef long long ll;
typedef double db;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
templ inline T rnd(T l,T r) {return uniform_int_distribution<T>(l,r)(rng);}
templ inline bool chkmax(T &x,T y){return x<y?x=y,1:0;}
templ inline bool chkmin(T &x,T y){return x>y?x=y,1:0;}
templ inline void read(T& t)
{
t=0;char f=0,ch=getchar();double d=0.1;
while(ch>'9'||ch<'0') f|=(ch=='-'),ch=getchar();
while(ch<='9'&&ch>='0') t=t*10+ch-48,ch=getchar();
if(ch=='.'){ch=getchar();while(ch<='9'&&ch>='0') t+=d*(ch^48),d*=0.1,ch=getchar();}
t=(f?-t:t);
}
template<typename T,typename... Args>inline void read(T& t,Args&... args){read(t); read(args...);}
char __sr[1<<21],__z[20];int __C=-1,__zz=0;
inline void Ot(){fwrite(__sr,1,__C+1,stdout),__C=-1;}
inline void print(register int x)
{
if(__C>1<<20)Ot();if(x<0)__sr[++__C]='-',x=-x;
while(__z[++__zz]=x%10+48,x/=10);
while(__sr[++__C]=__z[__zz],--__zz);__sr[++__C]='\n';
}
void file()
{
#ifdef NTFOrz
freopen("a.in","r",stdin);
#endif
}
inline void chktime()
{
#ifndef ONLINE_JUDGE
cout<<(clock()-t)/1000.0<<'\n';
#endif
}
ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x%mod) if (y&1) ret=ret*x%mod;return ret;}
// inline ll mul(ll a,ll b){ll d=(ll)(a*(double)b/mod+0.5);ll ret=a*b-d*mod;if (ret<0) ret+=mod;return ret;}
}
using namespace my_std;
int n,m,mm;
int inv[mod+5];
struct Int{short a,z;short v(){return z?0:a;}};
#define Int(x,y) ((Int){x,y})
Int operator + (Int a,Int b){return Int((a.v()+b.v())%mod,0);}
Int operator - (Int a,Int b){return Int((a.v()-b.v()+mod)%mod,0);};
Int operator * (Int a,Int b){ if (b.v()) a.a=1ll*a.a*b.a%mod; else a.z++; return a; }
Int operator / (Int a,Int b){ if (b.v()) a.a=1ll*a.a*inv[b.a]%mod; else a.z--; return a; }
struct Array
{
Int a[SS];
const Array operator + (const Array &x) const {Array ret;rep(i,0,m-1) ret.a[i]=a[i]+x.a[i];return ret;}
const Array operator - (const Array &x) const {Array ret;rep(i,0,m-1) ret.a[i]=a[i]-x.a[i];return ret;}
const Array operator * (const Array &x) const {Array ret;rep(i,0,m-1) ret.a[i]=a[i]*x.a[i];return ret;}
const Array operator / (const Array &x) const {Array ret;rep(i,0,m-1) ret.a[i]=a[i]/x.a[i];return ret;}
}fwt[SS];
struct Matrix
{
Array a,b,c,d;
const Matrix operator * (const Matrix &x) const {return (Matrix){a*x.a,a*x.b+b,c*x.a+x.c,c*x.b+d+x.d};}
};
void FWT(Array &a,int type)
{
Int p,q,I=Int(ksm(2,mod-2),0);
rep(i,0,mm-1)
for (int mid=1<<i,j=0;j<m;j+=mid<<1)
rep(k,0,mid-1)
{
p=a.a[j+k],q=a.a[j+k+mid];
if (type==1) a.a[j+k]=p+q,a.a[j+k+mid]=p-q;
else a.a[j+k]=(p+q)*I,a.a[j+k+mid]=(p-q)*I;
}
}
int val[sz];
struct hh{int t,nxt;}edge[sz<<1];
int head[sz],ecnt;
void make_edge(int f,int t)
{
edge[++ecnt]=(hh){t,head[f]};
head[f]=ecnt;
edge[++ecnt]=(hh){f,head[t]};
head[t]=ecnt;
}
int dfn[sz],pre[sz],size[sz],son[sz],top[sz],bot[sz],fa[sz],T;
#define v edge[i].t
void dfs1(int x,int f)
{
size[x]=1,fa[x]=f;
go(x) if (v!=f)
{
dfs1(v,x);
size[x]+=size[v];
if (size[v]>size[son[x]]) son[x]=v;
}
}
void dfs2(int x,int fa,int tp)
{
pre[dfn[bot[top[x]=tp]=x]=++T]=x;
if (son[x]) dfs2(son[x],x,tp);
go(x) if (v!=fa&&v!=son[x]) dfs2(v,x,v);
}
Array dp[sz],S[sz];
void dfs(int x,int fa)
{
dp[x]=S[x]=fwt[val[x]];
go(x) if (v!=fa)
{
dfs(v,x);
S[x]=S[x]+dp[x]*dp[v]+S[v];
dp[x]=dp[x]+dp[x]*dp[v];
}
}
#undef v
Matrix tr[sz<<2],tmp[sz];
#define lson k<<1,l,mid
#define rson k<<1|1,mid+1,r
void build(int k,int l,int r)
{
if (l==r)
{
int x=pre[l];Array f,s;f=s=fwt[val[x]];
#define v edge[i].t
go(x) if (v!=fa[x]&&v!=son[x]) s=s+f*dp[v]+S[v],f=f*(fwt[0]+dp[v]);
#undef v
tr[k]=tmp[l]=(Matrix){f,f,f,s};
return;
}
int mid=(l+r)>>1;
build(lson),build(rson);
tr[k]=tr[k<<1]*tr[k<<1|1];
}
void modify(int k,int l,int r,int x)
{
if (l==r) return (void)(tr[k]=tmp[l]);
int mid=(l+r)>>1;
if (x<=mid) modify(lson,x);
else modify(rson,x);
tr[k]=tr[k<<1]*tr[k<<1|1];
}
Matrix query(int k,int l,int r,int x,int y)
{
if (x<=l&&r<=y) return tr[k];
int mid=(l+r)>>1;
if (y<=mid) return query(lson,x,y);
if (x>mid) return query(rson,x,y);
return query(lson,x,y)*query(rson,x,y);
}
#undef lson
#undef rson
void modify(int x,int w)
{
Array p=tmp[dfn[x]].a,s=tmp[dfn[x]].d-p;
p=p/fwt[val[x]];p=p*fwt[w];val[x]=w;
tmp[dfn[x]].a=tmp[dfn[x]].b=tmp[dfn[x]].c=p;tmp[dfn[x]].d=s+p;
while (233)
{
Matrix a=query(1,1,n,dfn[top[x]],dfn[bot[top[x]]]);
modify(1,1,n,dfn[x]);
Matrix b=query(1,1,n,dfn[top[x]],dfn[bot[top[x]]]);
x=fa[top[x]]; if (!x) return;
Matrix &M=tmp[dfn[x]];
Array f0=M.a,s0=M.d;s0=s0-f0;
f0=f0/(fwt[0]+a.b);
f0=f0*(fwt[0]+b.b);
s0=s0-a.d+b.d;
tmp[dfn[x]]=(Matrix){f0,f0,f0,s0+f0};
}
}
int main()
{
file();
rep(i,1,mod-1) inv[i]=ksm(i,mod-2);
read(n,m);mm=log2(m);
rep(i,0,m-1) fwt[i].a[i]=Int(1,0),FWT(fwt[i],1);
rep(i,1,n) read(val[i]);
int x,y;
rep(i,1,n-1) read(x,y),make_edge(x,y);
dfs1(1,0),dfs2(1,0,1),dfs(1,0),build(1,1,n);
int Q;read(Q);char s[15];
while (Q--)
{
cin>>s;
if (s[0]=='C') read(x,y),modify(x,y);
else
{
read(x);
Array ans=query(1,1,n,1,dfn[bot[1]]).d;
FWT(ans,-1);
printf("%d\n",ans.a[x].v());
}
}
return 0;
}