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 N = 1e5 + 10;
const int logN = 20;
vector<int> G[N];
int in[N], out[N], L[N], P[N][logN], par[N];
vector<int> level[N];
int timer = 1;
void dfs(int u, int p, int l) {
  L[u] = l;
  P[u][0] = p;
  in[u] = timer++;
  level[l].push_back(in[u]);
  for (auto v : G[u]) {
    if (v == p) continue;
    dfs(v, u, l + 1);
  }
  out[u] = timer - 1;
}
int query(int u, int dist) {
  for (int i = logN - 1; i >= 0; i--)
    if (dist & (1 << i)) u = P[u][i];
  return u;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    if (!x) continue;
    G[x].push_back(i);
    G[i].push_back(x);
    par[i] = x;
  }
  for (int i = 1; i <= n; i++)
    if (par[i] == 0) dfs(i, 0, 0);
  for (int j = 1; j < logN; j++)
    for (int i = 1; i <= n; i++) P[i][j] = P[P[i][j - 1]][j - 1];
  int q;
  cin >> q;
  while (q--) {
    int u, h;
    cin >> u >> h;
    int v = query(u, h);
    if (!v) {
      cout << 0 << " ";
      continue;
    }
    int res = upper_bound(level[L[u]].begin(), level[L[u]].end(), out[v]) - 1 -
              lower_bound(level[L[u]].begin(), level[L[u]].end(), in[v]);
    cout << res << " ";
  }
  return 0;
}