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
#include <bits/stdc++.h>
using namespace std;
long long const delta = 1000000007;
long long par[(long long)1e5 + 8][18];
vector<long long> G[(long long)1e5 + 8];
vector<long long> height[(long long)1e5 + 8];
long long hi[(long long)1e5 + 8];
long long num[(long long)1e5 + 8];
long long sz[(long long)1e5 + 8];
long long ind = 1;
void lca(long long v) {
  for (long long i = 1; i < 18; i++) {
    long long u = par[v][i - 1];
    par[v][i] = par[u][i - 1];
    long long t = par[v][i];
  }
  for (long long i = 0; i < G[v].size(); i++) lca(G[v][i]);
}
void dfs(long long v, long long h) {
  num[v] = ind;
  height[h].push_back(num[v]);
  ;
  hi[v] = h;
  ind++;
  sz[num[v]] = 1;
  for (long long i = 0; i < G[v].size(); i++) {
    dfs(G[v][i], h + 1);
    sz[num[v]] += sz[num[G[v][i]]];
  }
}
long long ans(long long v, long long h) {
  if (!h) return 0;
  long long father = v;
  for (long long i = 17; i >= 0; i--) {
    if (h & (1 << i)) {
      father = par[father][i];
    }
  }
  if (!father) return 0;
  long long h1 = hi[father];
  long long h2 = h1 + h;
  long long n = num[father];
  long long lef, righ;
  long long l = 0, r = height[h2].size() - 1, mid = (l + r) / 2;
  while (r - l > 1) {
    if (height[h2][mid] < n)
      l = mid;
    else
      r = mid;
    mid = (l + r) / 2;
  }
  if (height[h2][l] >= n)
    lef = l;
  else
    lef = r;
  l = 0;
  r = height[h2].size() - 1;
  mid = (l + r) / 2;
  while (r - l > 1) {
    if (height[h2][mid] > n + sz[n] - 1)
      r = mid;
    else
      l = mid;
    mid = (l + r) / 2;
  }
  if (height[h2][r] <= n + sz[n] - 1)
    righ = r;
  else
    righ = l;
  return (righ - lef);
}
int main() {
  std::ios::sync_with_stdio(0);
  ;
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  long long t;
  for (long long i = 1; i <= n; i++) {
    cin >> t;
    par[i][0] = t;
    G[t].push_back(i);
    ;
  }
  for (long long i = 1; i <= n; i++) {
    if (!par[i][0]) {
      lca(i);
      dfs(i, 0);
    }
  }
  long long q;
  cin >> q;
  long long a, b;
  for (long long i = 0; i < q; i++) {
    cin >> a >> b;
    cout << ans(a, b) << endl;
  }
  return 0;
}
