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;
const int N = 1e5 + 7;
vector<int> g[N];
int cnt = 0, tin[N], tout[N], flor[N];
vector<int> flatin[N];
vector<int> flatout[N];
int up[N][25];
int len;
void dfs(int v, int p) {
  up[v][0] = p;
  tin[v] = ++cnt;
  flor[v] = flor[p] + 1;
  flatin[flor[v]].push_back(tin[v]);
  for (int i = 1; i <= len; i++) up[v][i] = up[up[v][i - 1]][i - 1];
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (to == p) continue;
    dfs(to, v);
  }
  tout[v] = ++cnt;
  flatout[flor[v]].push_back(tout[v]);
}
int bin_jump(int v, int p) {
  if (p == 1 || p == 0) {
    if (p == 1) {
      return up[v][0];
    } else {
      return v;
    }
  }
  int tw;
  tw = log2(p);
  return bin_jump(up[v][tw], p - (1 << tw));
}
int main() {
  int n;
  cin >> n;
  len = ceil(log2(n));
  for (int i = 1; i <= n; i++) {
    int a;
    scanf("%d", &a);
    if (!a) continue;
    g[a].push_back(i);
    g[i].push_back(a);
  }
  for (int i = 1; i <= n; i++) {
    if (!tin[i]) {
      dfs(i, 0);
    }
  }
  int m;
  cin >> m;
  while (m--) {
    int v, p;
    scanf("%d%d", &v, &p);
    int ppr = bin_jump(v, p);
    int flr;
    flr = flor[ppr] + p;
    if (!ppr) {
      cout << 0 << endl;
      continue;
    }
    int l, r;
    l = upper_bound(flatin[flr].begin(), flatin[flr].end(), tin[ppr]) -
        flatin[flr].begin();
    l++;
    r = lower_bound(flatout[flr].begin(), flatout[flr].end(), tout[ppr]) -
        flatout[flr].begin();
    int ans;
    ans = r - l;
    printf("%d\n", ans);
  }
}
