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, mxlg = 20, mod = 1e8, inf = 1e8;
int t, h[maxn], st[maxn], ft[maxn], par[mxlg][maxn];
vector<int> adj[maxn], height[maxn], stdfs;
void dfs(int v) {
  st[v] = t++;
  height[h[v]].push_back(st[v]);
  for (auto u : adj[v]) {
    h[u] = h[v] + 1;
    dfs(u);
  }
  ft[v] = t;
}
int find_par(int v, int h) {
  for (int i = 0; i < mxlg; i++)
    if ((h >> i) & 1) v = par[i][v];
  return v;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int u;
    cin >> u;
    if (u == 0) {
      par[0][i] = i;
      stdfs.push_back(i);
    } else {
      par[0][i] = --u;
      adj[u].push_back(i);
    }
  }
  for (int i = 1; i < mxlg; i++)
    for (int j = 0; j < n; j++) par[i][j] = par[i - 1][par[i - 1][j]];
  for (auto x : stdfs) {
    h[x] = 0;
    dfs(x);
  }
  int q;
  cin >> q;
  while (q--) {
    int v, p;
    cin >> v >> p;
    v--;
    if (h[v] < p)
      cout << "0 ";
    else {
      int u = find_par(v, p);
      cout << upper_bound(height[h[v]].begin(), height[h[v]].end(), ft[u]) -
                  lower_bound(height[h[v]].begin(), height[h[v]].end(), st[u]) -
                  1
           << " ";
    }
  }
  cout << "\n";
  return 0;
}