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 dep[maxn], fa[maxn][22];
int root[maxn], n, m, rk[maxn];
int cnt[maxn], block, belong[maxn], in[maxn], out[maxn], tim = 0;
vector<int> G[maxn];
void DFS(int u) {
  in[u] = ++tim;
  rk[tim] = u;
  for (int i = 0; i < G[u].size(); ++i) {
    int v = G[u][i];
    dep[v] = dep[u] + 1;
    fa[v][0] = u;
    for (int j = 1; (1 << j) <= n; ++j) fa[v][j] = fa[fa[v][j - 1]][j - 1];
    DFS(v);
  }
  out[u] = tim;
}
inline int seek(int x, int k) {
  for (int i = 0; (1 << i) <= k; ++i) {
    if (k & (1 << i)) x = fa[x][i];
  }
  return x;
}
inline void add(int x) { ++cnt[dep[x]]; }
inline void del(int x) { --cnt[dep[x]]; }
struct Node {
  int l, r, d, id;
  bool operator<(const Node& b) const {
    if (belong[l] == belong[b.l]) {
      return (belong[l] & 1) ? (r < b.r) : (r > b.r);
    } else {
      return belong[l] < belong[b.l];
    }
  }
} Q[maxn];
int ans[maxn];
int main() {
  scanf("%d", &n);
  block = sqrt(1.0 * n);
  root[0] = 0;
  for (int i = 1; i <= n; ++i) {
    belong[i] = (i - 1) / block + 1;
    int x;
    scanf("%d", &x);
    if (x)
      G[x].push_back(i);
    else {
      root[++root[0]] = i;
    }
  }
  for (int i = 1; i <= root[0]; ++i) {
    dep[root[i]] = 1;
    DFS(root[i]);
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; ++i) {
    int x, k;
    scanf("%d %d", &x, &k);
    int f = seek(x, k);
    Q[i] = (Node){in[f], out[f], dep[x], i};
  }
  sort(Q + 1, Q + m + 1);
  int L = 1, R = 0;
  for (int i = 1; i <= m; ++i) {
    while (R < Q[i].r) add(rk[++R]);
    while (R > Q[i].r) del(rk[R--]);
    while (L > Q[i].l) add(rk[--L]);
    while (L < Q[i].l) del(rk[L++]);
    ans[Q[i].id] = cnt[Q[i].d] - 1;
  }
  for (int i = 1; i <= m; ++i)
    printf("%d%c", max(0, ans[i]), i == m ? '\n' : ' ');
  return 0;
}
