The meaning of the question: Given a tree, each node on the tree has a corresponding character, ask multiple times whether any combination of characters on all nodes with a depth of \(d\) in the \(u\) subtree can make up a palindrome
Store the dfs order in a two-dimensional linear table, one dimension records the character and the other dimension records the depth.
Because the dfs order is monotonically increasing, the value of each two-dimensional table is also monotonically increasing,
so it is only necessary to use two binary divisions. my son can do it
It feels like a rather strange gesture, but I learned it anyway
//Is it really a man to be so violent in time and space?
#include<bits/stdc++.h>
#define rep(i,j,k) for(register int i=j;i<=k;i++)
#define rrep(i,j,k) for(register int i=j;i>=k;i--)
#define erep(i,u) for(register int i=head[u];~i;i=nxt[i])
#define print(a) printf("%lld",(ll)(a))
#define println(a) printf("%lld\n",(ll)(a))
#define printbk(a) printf("%lld ",(ll)(a))
using namespace std;
const int MAXN = 5e5+11;
typedef long long ll;
ll read(){
ll x=0,f=1;register char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int to[MAXN<<1],nxt[MAXN<<1],head[MAXN],tot;
int CLOCK,st[MAXN],ed[MAXN],n,m;
char str[MAXN];
vector<int> save[MAXN][27];
void init(){
memset(head,-1,sizeof head);
tot=CLOCK=0;
}
void add(int u,int v){
to[tot]=v;
nxt[tot]=head[u];
head[u]=tot++;
swap(u,v);
to[tot]=v;
nxt[tot]=head[u];
head[u]=tot++;
}
void dfs(int u,int p,int d){
st[u]=++CLOCK;
save[d][str[u]-'a'].push_back(st[u]);
erep(i,u){
int v=to[i];
if(v==p)continue;
dfs(v,u,d+1);
}
ed[u]=CLOCK;
}
int main(){
while(cin>>n>>m){
init();
rep(i,2,n) add(i,read());
scanf("%s",str+1);
memset(save,0,sizeof save);
dfs(1,-1,1);
rep(i,1,m){
int u=read();
int d=read();
bool flag=0,fflag=0;
rep(j,0,25){
vector<int>::iterator it1=lower_bound(save[d][j].begin(),save[d][j].end(),st[u]);
vector<int>::iterator it2=upper_bound(save[d][j].begin(),save[d][j].end(),ed[u]);
int cha=it2-it1;
if((cha&1)&&!flag) flag=1;
else if((cha&1)&&flag){
fflag=1;break;
}
}
if(fflag) printf("No\n");
else printf("Yes\n");
}
}
return 0;
}