Thinking
It is easy to think of \ (O (3 ^ {3m } \ log T) \) violence take great moment, not too obviously.
We analyze the nature of each transfer. Subject to the transfer equation is filling method, we try to change the brush table method to see ......
Discovery seems futile.
It noted that the rules of the game is to eat 1 0,2 1,0 eat eat 2, that is, \ (xy = 1 \ pmod 3 \) when \ (x \) eat \ (the y-\) .
We enumerate \ (J \) , and then enumerate \ (i \ ominus j \) (where subtraction is not every abdicated subtraction), according to \ (i \ ominus j \) to update the status of \ (f_i \ ) .
In other words, the enumeration \ (J, K \) , followed by \ (f_j \ times b_ {cnt1 [k], cnt2 [k]} \) to update \ (F_ {J \ K} Oplus \) .
This is found in the form of a high-dimensional circular convolution with FWT to do.
Due to the special properties of modulus, the modulus can be found in certain prime and 3, the final answer is obtained.
(Online solution to a problem so many do not understand why must forcibly analyze the nature of the transition matrix ah qwq)
Code
After just after I started to write ugly card is not in the past, learn minamoto gods code.
#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 550000
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
}
// 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,T,mod;
ll b[15][15];
int add(int x,int y)
{
int cur=1,ret=0;
rep(k,1,m) ret+=(x+y)%3*cur,cur*=3,x/=3,y/=3;
return ret;
}
struct hh{ll x,y;hh(ll X=0,ll Y=0){x=X,y=Y;}}f[sz],w[sz]; // x + y \times \omega
inline ll M1(ll x){return x-((mod-x-1)>>31&mod);}
inline ll M2(ll x){return x+(x>>31&mod);}
hh operator * (hh a,hh b) { return hh(M2(a.x*b.x%mod-a.y*b.y%mod),M2(M1(a.x*b.y%mod+a.y*b.x%mod)-a.y*b.y%mod)); }
hh operator + (hh a,hh b) { return hh(M1(a.x+b.x),M1(a.y+b.y)); }
bool operator < (hh a,hh b){return a.x==b.x?a.y<b.y:a.x<b.x;}
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;}
ll Ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x) if (y&1) ret=ret*x;return ret;}
hh kksm(hh x,int y){hh ret=hh(1,0);for (;y;y>>=1,x=x*x) if (y&1) ret=ret*x;return ret;}
void exgcd(ll a,ll b,ll &x,ll &y)
{
if (a==1&&!b) return (void)(x=1,y=0);
exgcd(b,a%b,y,x);y=M2(y-a/b*x%mod);
}
ll inv(ll x){ll a,b;exgcd(x,mod,a,b);return (a+mod)%mod;}
hh calc1(hh a){return hh(M2(-a.y),M2(a.x-a.y));}
hh calc2(hh a){return hh(M2(a.y-a.x),M2(-a.x));}
void FWT(hh *a,int type)
{
for (int len=1,i=0;i<m;i++,len*=3) for (int j=0;j<n;j+=len*3) rep(k,0,len-1)
{
hh x=a[j+k],y=a[j+k+len],z=a[j+k+len*2],Y1=calc1(y),Y2=calc2(y),Z1=calc1(z),Z2=calc2(z);
if (type==-1) swap(Y1,Y2),swap(Z1,Z2);
a[j+k]=x+y+z;
a[j+k+len]=x+Y1+Z2;
a[j+k+len+len]=x+Y2+Z1;
}
}
int main()
{
file();
read(m,T,mod);n=Ksm(3,m);
if (mod==1) { rep(i,0,n-1) puts("0"); return 0; }
rep(i,0,n-1) read(f[i].x);
rep(i,1,m+1) rep(j,1,m+2-i) read(b[i-1][j-1]);
rep(i,0,n-1)
{
int x=i,y=i,c1=0,c2=0;
rep(k,1,m) c1+=(x%3==1),x/=3;
rep(k,1,m) c2+=(y%3==2),y/=3;
w[i].x=b[c1][c2];
}
FWT(w,1);FWT(f,1);
map<hh,hh>M;
rep(i,0,n-1) w[i]=(M.count(w[i])?M[w[i]]:M[w[i]]=kksm(w[i],T))*f[i];
FWT(w,-1);
ll I=inv(n);
rep(i,0,n-1) printf("%lld\n",w[i].x*I%mod);
return 0;
}