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 depth hi. 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.
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 the i-th query.
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).
6 5 1 1 1 3 3 zacccd 1 1 3 3 4 1 6 1 1 2
Yes No Yes Yes Yes
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】给一棵树，树上各个节点对应一个字符，m个询问v，h，即v所在的子树节点中深度为h（在整棵树上的深度）的点的字符能否组成回文串。
【解题方法】dfs记录子树的出入时间，并将同一深度的同一字符保存在一起，对于输入的v以st[v]和en[v] 在对应的h深度不同字符段做二分查找，判断个数奇偶。但是这种方法是很慢的，我的代码跑了1900多ms险过，好像还有离线的做法，有空研究一下。
【AC 代码】
//
//Created by just_sort 2016/9/9 20:37
//Copyright (c) 2016 just_sort.All Rights Reserved
//
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 500010;
vector <int> ch[maxn];
vector <int> f[maxn][26];
int st[maxn],en[maxn],clk;
char s[maxn];
void dfs(int u,int d)
{
st[u]=++clk;
f[d][s[u-1]-'a'].push_back(clk);
for(int i=0; i<(int)ch[u].size(); i++)
dfs(ch[u][i],d+1);
en[u]=clk;
}
int main()
{
int n,m,v,h,x;
scanf("%d%d",&n,&m);
for(int i=2; i<=n; i++){
scanf("%d",&x);
ch[x].push_back(i);
}
scanf("%s",s);
clk = 0;
dfs(1,1);
while(m--)
{
scanf("%d%d",&v,&h);
int cnt=0;
for(int i=0; i<26; i++){
int temp = upper_bound(f[h][i].begin(),f[h][i].end(),en[v])-lower_bound(f[h][i].begin(),f[h][i].end(),st[v]);
if(temp%2) cnt++;
if(cnt>1) break;
}
if(cnt>1) printf("No\n");
else printf("Yes\n");
}
return 0;
}
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