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>
#pragma GCC optimize(2)
using namespace std;
struct node {
  int u, k;
};
int n, m;
int dp[100005][25], siz[100005];
int dep[100005], cnt[100005];
map<int, int> ans[100005];
vector<int> g[100005], p[100005], rt;
vector<node> ask;
void prepare_dfs(int u) {
  int maxx = -1, _ind = -1;
  for (int i = 1; i <= 20; i++) dp[u][i] = dp[dp[u][i - 1]][i - 1];
  siz[u] = 1;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    dep[v] = dep[u] + 1;
    dp[v][0] = u;
    prepare_dfs(v);
    siz[u] += siz[v];
    if (siz[v] > maxx) {
      maxx = siz[v];
      _ind = i;
    }
  }
  if (_ind == -1) return;
  swap(g[u][_ind], g[u][g[u].size() - 1]);
}
int jump(int u, int k) {
  int orik = k;
  for (int i = 20; i >= 0; i--) {
    if (k - (1 << i) >= 0) {
      k -= (1 << i);
      u = dp[u][i];
    }
  }
  p[u].push_back(orik);
  return u;
}
void fuckk(int u) {
  cnt[dep[u]]++;
  for (int i = 0; i < g[u].size(); i++) fuckk(g[u][i]);
}
void dfs(int u, int opt) {
  for (int i = 0; i < int(g[u].size()) - 1; i++) {
    int v = g[u][i];
    dfs(v, 0);
  }
  cnt[dep[u]]++;
  if (g[u].size()) dfs(g[u].back(), 1);
  for (int i = 0; i < int(g[u].size()) - 1; i++) fuckk(g[u][i]);
  for (int i = 0; i < p[u].size(); i++)
    ans[u][p[u][i]] = cnt[dep[u] + p[u][i]] - 1;
  if (opt == 0)
    for (int i = dep[u]; i <= n && cnt[i] != 0; i++) cnt[i] = 0;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1, fa; i <= n; i++) {
    cin >> fa;
    if (fa == 0)
      rt.push_back(i);
    else
      g[fa].push_back(i);
  }
  for (int i = 0; i < rt.size(); i++) prepare_dfs(rt[i]);
  cin >> m;
  for (int i = 1, U, K; i <= m; i++) {
    cin >> U >> K;
    ask.push_back((node){jump(U, K), K});
  }
  for (int i = 0; i < rt.size(); i++) {
    dfs(rt[i], 1);
    for (int j = dep[rt[i]]; j <= n && cnt[j]; j++) cnt[j] = 0;
  }
  for (int i = 0; i < ask.size(); i++) {
    if (ans[ask[i].u].count(ask[i].k) == 0) {
      cout << "0 ";
    } else {
      cout << ans[ask[i].u][ask[i].k] << ' ';
    }
  }
  return 0;
}
