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>
char rd;
int pn;
template <typename Type>
inline void read(Type &v) {
  pn = 1;
  while ((rd = getchar()) < '0' || rd > '9')
    if (rd == '-') pn = -1;
  v = rd - '0';
  while ((rd = getchar()) >= '0' && rd <= '9') v = v * 10 + rd - '0';
  v *= pn;
}
struct Edge {
  int v, next;
  Edge(int _v = 0, int nxt = 0) {
    v = _v;
    next = nxt;
  }
};
using namespace std;
const int maxn = 1e5 + 2;
int n, p[maxn][18], pre[maxn], post[maxn], dfs_clock, m, dep[maxn];
Edge edg[maxn];
int cnt, g[maxn];
vector<int> node[maxn];
void dfs(int u) {
  node[dep[u]].push_back(u);
  pre[u] = ++dfs_clock;
  Edge *e;
  for (int i = g[u]; i; i = e->next) {
    e = &edg[i];
    if (e->v != p[u][0]) {
      dep[e->v] = dep[u] + 1;
      dfs(e->v);
    }
  }
  post[u] = ++dfs_clock;
}
int kanc(int u, int k) {
  for (int j = 17; j >= 0; --j) {
    if ((1 << j) <= k) {
      u = p[u][j];
      k -= (1 << j);
    }
  }
  return u;
}
int sum(int u, int k) {
  int l, r, mid, v, dp = dep[u] + k, ans;
  l = 0;
  r = node[dp].size() - 1;
  while (l < r) {
    mid = (l + r) / 2;
    v = node[dp][mid];
    if (pre[v] < pre[u]) {
      l = mid + 1;
    } else
      r = mid;
  }
  ans = r;
  l = 0;
  r = node[dp].size() - 1;
  while (l < r) {
    mid = (l + r) / 2 + 1;
    v = node[dp][mid];
    if (post[v] <= post[u])
      l = mid;
    else
      r = mid - 1;
  }
  return r - ans;
}
int main() {
  read(n);
  int x, y, z;
  for (int i = 1; i <= n; ++i) {
    read(x);
    p[i][0] = x;
    edg[++cnt] = Edge(i, g[x]);
    g[x] = cnt;
    ;
  }
  dfs(0);
  for (int j = 1; j < 18; ++j) {
    for (int i = 1; i <= n; ++i) {
      p[i][j] = p[p[i][j - 1]][j - 1];
    }
  }
  read(m);
  for (int i = 1; i <= m; ++i) {
    read(x);
    read(y);
    z = kanc(x, y);
    if (z)
      printf("%d ", sum(z, y));
    else
      printf("0 ");
  }
}
