Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize(3)
#pragma GCC optimize(2)
using namespace std;
const int maxn = 1e5 + 10;
vector<int> g[maxn];
vector<int>* vec[maxn];
vector<pair<int, int> > qs[maxn];
int depth[maxn];
int cnt[maxn];
int ans[maxn];
int sz[maxn];
int up[maxn][25];
int N, Q, L;
bool root[maxn];
void getsz(int v, int p, int d) {
  up[v][0] = p;
  sz[v] = 1;
  depth[v] = d;
  for (int u : g[v]) {
    if (u != p) {
      getsz(u, v, d + 1);
      sz[v] += sz[u];
    }
  }
  if (0) {
    printf("sz[%d] = %d\n", v, sz[v]);
    printf("depth[%d] = %d\n", v, depth[v]);
  }
}
void add(int v, int first) {
  vec[v]->push_back(first);
  ++cnt[depth[first]];
}
void DFS(int v, int p, bool keep) {
  int mx = -1, bigChild = -1;
  for (int u : g[v]) {
    if (u != p && mx < sz[u]) {
      mx = sz[u];
      bigChild = u;
    }
  }
  for (int u : g[v]) {
    if (u != p && u != bigChild) {
      DFS(u, v, 0);
    }
  }
  if (bigChild != -1) {
    DFS(bigChild, v, 1);
    vec[v] = vec[bigChild];
  } else {
    vec[v] = new vector<int>();
  }
  add(v, v);
  for (int u : g[v]) {
    if (u != p && u != bigChild) {
      for (int first : *vec[u]) {
        add(v, first);
      }
    }
  }
  for (auto& e : qs[v]) {
    ans[e.second] = cnt[e.first] - 1;
  }
  if (0) {
    printf("at tree rooted at %d\n", v);
    for (int i = 0; i <= N; ++i) {
      printf("cnt[%d] = %d\n", i, cnt[i]);
    }
    printf("\n");
  }
  if (keep == 0) {
    for (int i : *vec[v]) {
      --cnt[depth[i]];
    }
  }
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; ++i) {
    int first;
    scanf("%d", &first);
    if (first == 0) {
      root[i] = 1;
    } else {
      g[i].push_back(first);
      g[first].push_back(i);
    }
  }
  for (int i = 1; i <= N; ++i) {
    if (root[i]) {
      getsz(i, i, 1);
    }
  }
  L = ceil(log2(N));
  for (int j = 1; j <= L; ++j) {
    for (int i = 1; i <= N; ++i) {
      up[i][j] = up[up[i][j - 1]][j - 1];
    }
  }
  if (0) {
    for (int j = 0; j <= L; ++j) {
      for (int i = 1; i <= N; ++i) {
        printf("up[%d][%d] = %d\n", i, j, up[i][j]);
      }
    }
  }
  scanf("%d", &Q);
  for (int q = 1; q <= Q; ++q) {
    int v, h;
    scanf("%d %d", &v, &h);
    if (depth[v] < h + 1) continue;
    int vv = v;
    for (int i = L; i >= 0; --i) {
      if ((1 << i) <= h) {
        v = up[v][i];
        h -= (1 << i);
      }
    }
    if (0) printf("ask (%d %d) at %d\n", vv, q, v);
    qs[v].emplace_back(depth[vv], q);
  }
  for (int i = 1; i <= N; ++i) {
    if (root[i]) {
      DFS(i, i, 0);
    }
  }
  for (int i = 1; i <= Q; ++i) {
    printf("%d ", ans[i]);
  }
  printf("\n");
}