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;
const int maxn = 1e5 + 10;
int par[maxn][22];
int f[maxn];
int h[maxn];
int a[maxn];
vector<pair<int, int> > q[maxn];
vector<int> ans;
vector<int> adj[maxn];
int parent(int u, int p) {
  for (int i = 0; i < 21; i++) {
    if ((1 << i) & (p)) {
      u = par[u][i];
      if (u == -1) {
        return -1;
      }
    }
  }
  return u;
}
void dfs(int u) {
  if (h[u] != -1) a[h[u]]++;
  for (auto z : q[u]) {
    ans[z.first] = a[z.second + h[u]];
  }
  for (auto v : adj[u]) {
    h[v] = h[u] + 1;
    dfs(v);
  }
  for (auto z : q[u]) {
    ans[z.first] = a[z.second + h[u]] - ans[z.first];
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  memset(par, -1, sizeof par);
  for (int i = 0; i < n; i++) {
    int t;
    cin >> t;
    t--;
    f[i] = t;
    par[i][0] = t;
    if (t > -1) {
      adj[t].push_back(i);
    } else {
      adj[n + 1].push_back(i);
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 1; par[i][j - 1] != -1; j++) {
      par[i][j] = par[par[i][j - 1]][j - 1];
    }
  }
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int v, p;
    cin >> v >> p;
    v--;
    int x = parent(v, p);
    if (x != -1) {
      q[x].push_back({i, p});
    }
    ans.push_back(0);
  }
  h[n + 1] - 1;
  dfs(n + 1);
  for (auto i : ans) {
    if (i == 0) i++;
    cout << i - 1 << ' ';
  }
}