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 maxn4 = 1e4 + 85, maxn5 = 1e5 + 69, maxn6 = 1e6 + 85,
          maxn7 = 1e7 + 69, maxn8 = 1e8 + 85, maxn9 = 1e9 + 69, modi1 = 1e7 + 7,
          modi2 = 1e9 + 7;
const long long int mod1 = 1e7 + 7, mod2 = 1e9 + 7;
int mn = INT_MAX, mx = INT_MIN;
int n, par[16][maxn5], h[maxn5], sz[maxn5], st[maxn5], ft[maxn5], rime;
vector<int> g[maxn5];
vector<int> height[maxn5];
bool mark[maxn5];
void dfs(int s, int parr) {
  h[s] = h[parr] + 1;
  st[s] = rime++;
  mark[s] = 1;
  height[h[s]].push_back(st[s]);
  for (auto u : g[s]) {
    if (u == parr) continue;
    par[0][u] = s;
    dfs(u, s);
  }
  ft[s] = rime;
}
int get_kth_par(int v, int k) {
  if (k > h[v]) return 0;
  for (int i = 0; i < 16 and k; i++)
    if (k & (1 << i)) v = par[i][v], k ^= (1 << i);
  return v;
}
int ans(int p, int k) {
  k += h[p];
  return lower_bound((height[k]).begin(), (height[k]).end(), ft[p]) -
         lower_bound((height[k]).begin(), (height[k]).end(), st[p]);
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    if (x) g[x].push_back(i), g[i].push_back(x);
  }
  for (int i = 1; i <= n; i++)
    if (!mark[i]) dfs(i, 0);
  for (int i = 1; i < 16; i++)
    for (int j = 1; j <= n; j++) par[i][j] = par[i - 1][par[i - 1][j]];
  int q;
  cin >> q;
  while (q--) {
    int v, k;
    cin >> v >> k;
    int p = get_kth_par(v, k);
    if (p == 0) {
      cout << "0 ";
      continue;
    }
    cout << ans(p, k) - 1 << ' ';
  }
  return 0;
}