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 long long N = 1e5 + 5;
long long block;
vector<long long> G[N], adj[N];
long long L[N], dis[N], fin[N], timer = 1, p[25][N], T[N], logs[N];
long long arr[N], n, ans[N], freq[N], nodes[N];
bool vis[N];
struct Query {
  long long L, R, x, idx;
};
Query Q[100005];
bool cmp(Query a, Query b) {
  if (a.L / block != b.L / block) {
    return a.L < b.L;
  }
  return a.R < b.R;
}
void comp_logs() {
  logs[0] = 1;
  for (long long i = 1; i < N; i++) {
    logs[i] = logs[i / 2] + 1;
  }
}
void euler_tour(long long u, long long par, long long dep) {
  L[u] = dep;
  dis[u] = timer;
  arr[timer] = dep;
  timer++;
  T[u] = par;
  for (long long v : G[u]) {
    euler_tour(v, u, dep + 1);
  }
  fin[u] = timer - 1;
}
void build_lca() {
  memset(p, -1, sizeof(p));
  long long i, j;
  for (i = 1; i <= n; i++) {
    p[0][i] = T[i];
  }
  for (i = 1; i <= logs[n]; i++) {
    for (j = 1; j <= n; j++) {
      if (p[i - 1][j] != -1) {
        p[i][j] = p[i - 1][p[i - 1][j]];
      }
    }
  }
}
long long find_kth_par(long long u, long long k) {
  long long i, j;
  for (i = logs[k]; i >= 0; i--) {
    if (k - (1 << i) >= 0) {
      k -= (1 << i);
      u = p[i][u];
    }
  }
  return u;
}
void add(long long id) { freq[arr[id]]++; }
void rem(long long id) { freq[arr[id]]--; }
int main() {
  comp_logs();
  long long i, u, m, q, k, kth, pos = 0;
  scanf("%lld", &n);
  for (i = 1; i <= n; i++) {
    scanf("%lld", &u);
    nodes[i] = u;
    G[u].push_back(i);
  }
  for (i = 1; i <= n; i++) {
    if (!vis[i] && nodes[i] == 0) {
      euler_tour(i, -1, 0);
    }
  }
  build_lca();
  scanf("%lld", &m);
  q = m;
  for (i = 1; i <= m; i++) {
    scanf("%lld %lld", &u, &k);
    if (L[u] < k) {
      Q[i].x = -1;
      Q[i].idx = i;
      continue;
    }
    kth = find_kth_par(u, k);
    Q[i].L = dis[kth];
    Q[i].R = fin[kth];
    Q[i].x = L[kth] + k;
    Q[i].idx = i;
  }
  block = sqrt(timer) + 2;
  sort(Q, Q + m, cmp);
  long long ml = 1, mr = 0;
  for (i = 1; i <= m; i++) {
    long long l = Q[i].L;
    long long r = Q[i].R;
    if (Q[i].x == -1) {
      ans[Q[i].idx] = 0;
    } else {
      while (ml > l) {
        ml--;
        add(ml);
      }
      while (mr < r) {
        mr++;
        add(mr);
      }
      while (ml < l) {
        rem(ml);
        ml++;
      }
      while (mr > r) {
        rem(mr);
        mr--;
      }
      ans[Q[i].idx] = freq[Q[i].x] - 1;
    }
  }
  for (i = 1; i <= m; i++) printf("%lld ", ans[i]);
  printf("\n");
}
