Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<vector<int> > g;
const int N = 1e5 + 5;
int n, p[N], tim = 1, in[N], out[N], lvl[N], sparse[N][20];
vector<pair<int, int> > group[N];
void dfs(int node) {
  in[node] = tim++;
  lvl[node] = lvl[p[node]] + 1;
  group[lvl[node]].push_back(make_pair(in[node], node));
  for (int i = 0; i < g[node].size(); i++) dfs(g[node][i]);
  out[node] = tim++;
}
void build() {
  for (int i = 1; i <= 18; i++) {
    for (int j = 1; j <= n; j++) {
      if (sparse[j][i - 1] != -1)
        sparse[j][i] = sparse[sparse[j][i - 1]][i - 1];
    }
  }
}
int solve(int v, int p) {
  int pthparent = v;
  for (int i = 0; i <= 18; i++) {
    if (p & (1 << i)) {
      pthparent = sparse[pthparent][i];
    }
    if (pthparent <= 0) return 0;
  }
  int tin = in[pthparent], tout = out[pthparent];
  int last = lower_bound(group[lvl[v]].begin(), group[lvl[v]].end(),
                         make_pair(tout, 0)) -
             group[lvl[v]].begin();
  int first = lower_bound(group[lvl[v]].begin(), group[lvl[v]].end(),
                          make_pair(tin, 0)) -
              group[lvl[v]].begin();
  return last - first - 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  memset(sparse, -1, sizeof sparse);
  cin >> n;
  g.resize(n + 1);
  for (int i = 1; i <= n; i++) {
    cin >> p[i];
    sparse[i][0] = p[i];
    g[p[i]].push_back(i);
  }
  dfs(0);
  build();
  int q;
  cin >> q;
  while (q--) {
    int v, p;
    cin >> v >> p;
    cout << solve(v, p) << ' ';
  }
  return 0;
}