描述

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 p**i, the parent index is always less than the index of the vertex (i.e., p**i

 < 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 v**i, h**i. Let’s consider the vertices in the subtree v**i located at depth h**i. 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 p*2, *p*3, …, *p**n — the parents of vertices from the second to the n-th (1 ≤ p**i < 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 v**i, h**i (1 ≤ v**i, h**i ≤ n) — the vertex and the depth that appear in the i-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).

input

6 5
1 1 1 3 3
zacccd
1 1
3 3
4 1
6 1
1 2

output

Yes
No
Yes
Yes
Yes

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”.

思路

題目給了一個n$n$個節點的樹,每一個節點上都有一個字母。然後有m$m$組查詢，每次查詢v h，代表查詢以v為根的子樹，在原來的樹中深度為h的點上面的字母能否構成一個迴文串。

首先能夠成一個迴文串的條件是: 最多隻有一個字母出現奇數次

首先我們對這一棵樹按照dfs序進行深搜，儲存一下，進入這個點的時間戳l[i]和離開這個點的時間戳r[i]，這樣當一個點的時間戳>=l[i] && <=r[i]就說明這個點是原來的點的子節點。

在dfs的時候用vector儲存一下當前這一層的每一個字母出現的點的時間戳l[i].

最後查詢的時候遍歷一下每一個字母二分找到這一層中，>=l[x] && <=r[x]的個數，這樣就知道滿足條件的字母出現了多少次，然後我們利用最多隻有一個字母出現奇數次的性質來判斷是可以構成迴文串。

程式碼

#include <bits/stdc++.h>
using namespace std;
typedef pair<int, int> pir;
#define mem(a, b) memset(a, b, sizeof(a))
const int N = 5e5 + 10;
vector<int> e[N];
vector<int> dp[N][27];
int l[N], r[N], id, dep[N];
char s[N];
int n, m;
void dfs(int u, int deep)
{
    dep[u] = deep;
    l[u] = ++id;
    dp[deep][s[u]].push_back(id);
    for (auto v : e[u])
        dfs(v, deep + 1);
    r[u] = id;
}
int main()
{
    //freopen("in.txt", "r", stdin);
    int x;
    scanf("%d%d", &n, &m);
    for (int i = 2; i <= n; i++)
    {
        scanf("%d", &x);
        e[x].push_back(i);
    }
    scanf("%s", s + 1);
    for (int i = 1; i <= n; i++)
        s[i] = s[i] - 'a';
    dfs(1, 1);
    int h;
    while (m--)
    {
        scanf("%d%d", &x, &h);
        int sum = 0;
        for (int i = 0; i < 26; i++)
        {
            int num = upper_bound(dp[h][i].begin(), dp[h][i].end(), r[x]) - lower_bound(dp[h][i].begin(), dp[h][i].end(), l[x]);
            if (num & 1)
            {
                sum++;
                if (sum > 1)
                    break;
            }
        }
        if (sum > 1)
            puts("No");
        else
            puts("Yes");
    }
    return 0;
}