Polycarpus got hold of a family tree. The found tree describes the family relations of n people, numbered from 1 to n. Every person in this tree has at most one direct ancestor. Also, each person in the tree has a name, the names are not necessarily unique.

We call the man with a number a a 1-ancestor of the man with a number b, if the man with a number a is a direct ancestor of the man with a number b.

We call the man with a number a a k-ancestor (k > 1) of the man with a number b, if the man with a number b has a 1-ancestor, and the man with a number a is a (k - 1)-ancestor of the 1-ancestor of the man with a number b.

In the tree the family ties do not form cycles. In other words there isn't a person who is his own direct or indirect ancestor (that is, who is an x-ancestor of himself, for some x, x > 0).

We call a man with a number a the k-son of the man with a number b, if the man with a number b is a k-ancestor of the man with a number a.

Polycarpus is very much interested in how many sons and which sons each person has. He took a piece of paper and wrote m pairs of numbers vi, ki. Help him to learn for each pair vi, ki the number of distinct names among all names of the ki-sons of the man with number vi.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 105) — the number of people in the tree. Next n lines contain the description of people in the tree. The i-th line contains space-separated string si and integer ri (0 ≤ ri ≤ n), where si is the name of the man with a number i, and ri is either the number of the direct ancestor of the man with a number i or 0, if the man with a number i has no direct ancestor. 

The next line contains a single integer m (1 ≤ m ≤ 105) — the number of Polycarpus's records. Next m lines contain space-separated pairs of integers. The i-th line contains integers vi, ki (1 ≤ vi, ki ≤ n).

It is guaranteed that the family relationships do not form cycles. The names of all people are non-empty strings, consisting of no more than 20 lowercase English letters.

Output

Print m whitespace-separated integers — the answers to Polycarpus's records. Print the answers to the records in the order, in which the records occur in the input.

Examples

Input

6
pasha 0
gerald 1
gerald 1
valera 2
igor 3
olesya 1
5
1 1
1 2
1 3
3 1
6 1


Output

2
2
0
1
0


Input

6
valera 0
valera 1
valera 1
gerald 0
valera 4
kolya 4
7
1 1
1 2
2 1
2 2
4 1
5 1
6 1


Output

1
0
0
0
2
0
0
#include <bits/stdc++.h>
using namespace std;
int deep[100005];
int val[300005];
int nxt[300005];
int cnt = 0;
int dsp[100005];
int dep[100005];
int ans[100005];
int h1[100005];
int h2[100005];
int h3[100005];
inline void add(int &H, int v) {
  nxt[++cnt] = H;
  val[cnt] = v;
  H = cnt;
}
int idf[200005];
int pnt = 0;
void dfs(int now) {
  idf[++pnt] = -now;
  for (register int v = h1[now]; v; v = nxt[v]) dfs(val[v]);
  idf[++pnt] = now;
}
void dfs1(int x) {
  for (register int i = 1; i <= pnt; i++)
    if (idf[i] > 0) {
      int now = idf[i];
      for (register int v = h2[now]; v; v = nxt[v])
        if (dsp[val[v]] < deep[now]) {
          add(h3[deep[now] - dsp[val[v]]], val[v]);
        }
      h2[now] = h3[deep[now]];
      h3[deep[now]] = 0;
    }
}
void dfs2(int now) {
  for (register int i = 1; i <= pnt; i++) {
    int now = abs(idf[i]), st = idf[i] >= 0 ? 1 : -1;
    if (idf[i] < 0) dep[deep[now]]++;
    for (register int v = h2[now]; v; v = nxt[v]) {
      ans[val[v]] += dep[deep[now] + dsp[val[v]]] * st;
    }
  }
}
int main() {
  int n, m;
  cin >> n;
  for (int i = 2; i <= n + 1; i++) {
    static int v;
    scanf("%d", &v);
    v++;
    deep[i] = deep[v] + 1;
    add(h1[v], i);
  }
  cin >> m;
  for (int i = 1; i <= m; i++) {
    static int u, v;
    scanf("%d%d", &u, &v);
    u++;
    add(h2[u], i);
    dsp[i] = v;
  }
  dfs(1);
  dfs1(1);
  dfs2(1);
  for (int i = 1; i <= m; i++) {
    printf("%d ", ans[i] - (bool)ans[i]);
  }
  return 0;
}
