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;
using ll = long long;
using ld = long double;
int node[(ll)(1e5 + 7)], in[(ll)(1e5 + 7)], out[(ll)(1e5 + 7)],
    ti = 1, hie[(ll)(1e5 + 7)];
int dp[(ll)(1e5 + 7)][17];
vector<int> gr[(ll)(1e5 + 7)], hi[(ll)(1e5 + 7)];
void dfs(int node, int h) {
  hi[h].push_back(ti);
  hie[node] = h;
  in[node] = ti++;
  for (int x : gr[node]) {
    dfs(x, h + 1);
  }
  out[node] = ti++;
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  vector<int> ve;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 17; j++) {
      dp[i][j] = -1;
    }
  }
  for (int i = 1; i <= n; i++) {
    int val;
    cin >> val;
    if (val) {
      dp[i][0] = val;
      gr[val].push_back(i);
    } else {
      ve.push_back(i);
    }
  }
  for (int x : ve) {
    dfs(x, 1);
  }
  for (int j = 1; j < 17; j++) {
    for (int i = 1; i <= n; i++) {
      if (dp[i][j - 1] != -1) {
        dp[i][j] = dp[dp[i][j - 1]][j - 1];
      }
    }
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int x, p;
    cin >> x >> p;
    for (int j = 0; (1 << j) <= p; j++) {
      if ((1 << j) & p) {
        x = dp[x][j];
      }
    }
    if (x == -1) {
      cout << "0 ";
      continue;
    }
    int hh = hie[x] + p;
    int xx = (upper_bound(hi[hh].begin(), hi[hh].end(), out[x]) -
              lower_bound(hi[hh].begin(), hi[hh].end(), in[x]));
    cout << max(0, xx - 1) << " ";
  }
  cout << "\n";
}
