codeforce 570D Tree Requests(dfs+位运算,好题)

D. Tree Requests

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Roman planted a tree consisting of n vertices. Each vertex contains a lowercase English letter. Vertex 1 is the root of the tree, each of the n - 1 remaining vertices has a parent in the tree. Vertex is connected with its parent by an edge. The parent of vertex i is vertex pi, the parent index is always less than the index of the vertex (i.e., pi < i).

The depth of the vertex is the number of nodes on the path from the root to v along the edges. In particular, the depth of the root is equal to 1.

We say that vertex u is in the subtree of vertex v, if we can get from u to v, moving from the vertex to the parent. In particular, vertex v is in its subtree.

Roma gives you m queries, the i-th of which consists of two numbers vi, hi. Let's consider the vertices in the subtree vi located at depthhi. Determine whether you can use the letters written at these vertices to make a string that is a palindrome. The letters that are written in the vertexes, can be rearranged in any order to make a palindrome, but all letters should be used.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 500 000) — the number of nodes in the tree and queries, respectively.

The following line contains n - 1 integers p2, p3, ..., pn — the parents of vertices from the second to the n-th (1 ≤ pi < i).

The next line contains n lowercase English letters, the i-th of these letters is written on vertex i.

Next m lines describe the queries, the i-th line contains two numbers vi, hi (1 ≤ vi, hi ≤ n) — the vertex and the depth that appear in thei-th query.

Output

Print m lines. In the i-th line print "Yes" (without the quotes), if in the i-th query you can make a palindrome from the letters written on the vertices, otherwise print "No" (without the quotes).

Examples

input

6 51 1 1 3 3zacccd1 13 34 16 11 2

output

YesNoYesYesYes

Note

String s is a palindrome if reads the same from left to right and from right to left. In particular, an empty string is a palindrome.

Clarification for the sample test.

In the first query there exists only a vertex 1 satisfying all the conditions, we can form a palindrome "z".

In the second query vertices 5 and 6 satisfy condititions, they contain letters "с" and "d" respectively. It is impossible to form a palindrome of them.

In the third query there exist no vertices at depth 1 and in subtree of 4. We may form an empty palindrome.

In the fourth query there exist no vertices in subtree of 6 at depth 1. We may form an empty palindrome.

In the fifth query there vertices 2, 3 and 4 satisfying all conditions above, they contain letters "a", "c" and "c". We may form a palindrome "cac".

题意：参考51nod翻译

题解：参考博客

由于这个是离线算法，与处理啊每个结点需要的询问，在dfs访问到该节点时一起处理，因为dfs该结点子树之前子树的信息还不在st中，dfs完后信息已经在st中，所以这两个相异或就是子树的信息了。

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<vector>using namespace std;#define rep(i,a,n) for (int i=a;i<n;i++)#define per(i,a,n) for (int i=n-1;i>=a;i--)#define pb push_back#define mp make_pair#define fi first#define se secondtypedef vector<int> VI;typedef long long ll;typedef pair<int,int> PII;const ll mod=1000000007;const int MAXN=500000+100;struct node{int to,next;}edge[MAXN*2];int tol=0;int head[MAXN];void add(int u,int v){edge[++tol].to=v,edge[tol].next=head[u],head[u]=tol;edge[++tol].to=u,edge[tol].next=head[v],head[v]=tol;}vector<PII> query[MAXN];int ans[MAXN],st[MAXN];char s[MAXN];void dfs(int x,int fa,int d){st[d]^=1<<(s[x]-'a');        for(int i=0;i<query[x].size();i++)  //先记录dfs之前的状态 {PII t=query[x][i];int j=t.fi,h=t.se;ans[j]^=st[h];}for(int i=head[x];i;i=edge[i].next){int v=edge[i].to;if(v==fa) continue;dfs(v,x,d+1);}for(int i=0;i<query[x].size();i++)  //与dfs后的状态比较,就可以知道该子树下的状态 {PII t=query[x][i];int j=t.fi,h=t.se;ans[j]^=st[h];}}int main(){int n,m;scanf("%d%d",&n,&m);rep(i,2,n+1){int v;scanf("%d",&v);add(i,v);}scanf("%s",s+1);rep(i,0,m){int v,h;scanf("%d%d",&v,&h);query[v].pb(mp(i,h));}dfs(1,-1,1);rep(i,0,m){if(ans[i]&(ans[i]-1)) puts("No");else puts("Yes");}return 0;}