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;
vector<int> adj[100010];
vector<pair<int, int>> q[100010];
int cnt[100010], sz[100010], dep[100010], ans[100010], par[100010][26];
bool big[100010];
void pre(int n) {
  for (int j = 1; j <= 20; ++j) {
    for (int i = 1; i <= n; ++i) {
      if (par[i][j - 1] != -1) par[i][j] = par[par[i][j - 1]][j - 1];
    }
  }
}
void dfs(int u, int p) {
  sz[u] = 1;
  dep[u] = dep[p] + 1;
  for (auto v : adj[u]) {
    if (v == p) continue;
    if (u) par[v][0] = u;
    dfs(v, u);
    sz[u] += sz[v];
  }
}
void add(int u, int p, int x) {
  cnt[dep[u]] += x;
  for (auto v : adj[u]) {
    if (v != p && !big[v]) {
      add(v, u, x);
    }
  }
}
int klevpar(int u, int k) {
  for (int i = 20; i >= 0; --i) {
    if (k & (1 << i)) {
      u = par[u][i];
      if (u == -1) break;
    }
  }
  return u;
}
void dsu(int u, int p, bool keep) {
  int bigchild = -1, mx = -1;
  for (auto v : adj[u]) {
    if (v != p && sz[v] > mx) {
      mx = sz[v];
      bigchild = v;
    }
  }
  for (auto v : adj[u]) {
    if (v != p && v != bigchild) {
      dsu(v, u, 0);
    }
  }
  if (bigchild != -1) {
    dsu(bigchild, u, 1);
    big[bigchild] = 1;
  }
  add(u, p, 1);
  for (auto x : q[u]) {
    ans[x.second] = cnt[dep[u] + x.first] - 1;
  }
  if (bigchild != -1) {
    big[bigchild] = 0;
  }
  if (keep == 0) {
    add(u, p, -1);
  }
}
int main() {
  int n, m;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int r;
    scanf("%d", &r);
    adj[r].push_back(i);
  }
  memset(par, -1, sizeof par);
  dfs(0, 0);
  pre(n);
  scanf("%d", &m);
  for (int i = 0; i < m; ++i) {
    int v, k;
    scanf("%d %d", &v, &k);
    v = klevpar(v, k);
    if (v != -1)
      q[v].emplace_back(k, i);
    else
      ans[i] = 0;
  }
  dsu(0, 0, 1);
  for (int i = 0; i < m; ++i) {
    printf("%d ", ans[i]);
  }
  printf("\n");
}
