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;
const int MAXN = 1e5 + 7;
int m, n;
int r[MAXN];
vector<int> edges[MAXN];
vector<int> dep[MAXN];
int Left[MAXN], Right[MAXN];
int d[MAXN];
void Input() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> r[i];
}
int t[MAXN][18];
int cnt2;
void DFS(int u, int cnt) {
  Left[u] = ++cnt2;
  d[u] = cnt;
  dep[cnt].push_back(Left[u]);
  for (auto v : edges[u])
    if (!d[v]) {
      t[v][0] = u;
      DFS(v, cnt + 1);
    }
  Right[u] = cnt2;
}
void Add(int u, int v) { edges[u].push_back(v); }
void Init() {
  for (int i = 1; i <= n; i++)
    if (r[i]) {
      Add(i, r[i]);
      Add(r[i], i);
    }
  for (int i = 1; i <= n; i++)
    if (!r[i]) DFS(i, 1);
  for (int i = 1; i <= 17; i++)
    for (int j = 1; j <= n; j++) t[j][i] = t[t[j][i - 1]][i - 1];
  for (int i = 1; i <= n; i++) sort(dep[i].begin(), dep[i].end());
}
int LCA(int v, int p) {
  for (int i = 0; i <= 16; i++)
    if (p & (1 << i)) {
      v = t[v][i];
      p -= (1 << i);
      if (!p) break;
    }
  return v;
}
int Result(int x, int p) {
  if (x == 0) return 0;
  int cnt = d[x] + p;
  if ((int)(dep[cnt]).size() - 1 == -1) return 0;
  int l = 0, r = (int)(dep[cnt]).size() - 1;
  int min_Left = r + 1;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (dep[cnt][mid] >= Left[x]) {
      min_Left = min(min_Left, mid);
      r = mid - 1;
    } else
      l = mid + 1;
  }
  l = min_Left, r = (int)(dep[cnt]).size() - 1;
  int max_Right = l - 1;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (dep[cnt][mid] <= Right[x]) {
      max_Right = max(max_Right, mid);
      l = mid + 1;
    } else
      r = mid - 1;
  }
  return max(0, max_Right - min_Left);
}
void Output() {
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int v, p;
    cin >> v >> p;
    cout << Result(LCA(v, p), p) << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  Input();
  Init();
  Output();
  return 0;
}
