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>
using namespace std;
const int INF = 0x3f3f3f3f;
const double EPS = 1e-9;
int n, son[100015], cnt[100015], deep[100015], h[100015];
vector<int> e[100015];
int pa[100015][20];
void dfs1(int u) {
  for (auto v : e[u]) {
    pa[v][0] = u;
    h[v] = h[u] + 1;
    dfs1(v);
    deep[u] = max(deep[u], deep[v]);
    if (0 == son[u] || deep[son[u]] < deep[v]) {
      son[u] = v;
    }
  }
  deep[u] += 1;
}
void pre() {
  pa[0][0] = -1;
  for (int d = 1; d <= 18; ++d) {
    for (int i = 0; i <= n; ++i) {
      int anc = pa[i][d - 1];
      if (-1 == anc) {
        pa[i][d] = -1;
      } else {
        pa[i][d] = pa[anc][d - 1];
      }
    }
  }
}
int query_pth_anc(int u, int p) {
  for (int i = 0; i <= 18; ++i) {
    if (p & 1) {
      u = pa[u][i];
    }
    p >>= 1;
    if (-1 == u) break;
  }
  return u;
}
map<int, long long> query[100015];
vector<int> *vec[100015];
void dfs2(int u) {
  for (auto v : e[u]) {
    if (son[u] == v) continue;
    dfs2(v);
  }
  if (son[u]) {
    dfs2(son[u]);
    vec[u] = vec[son[u]];
  } else {
    vec[u] = new vector<int>();
  }
  int sizu = vec[u]->size();
  for (auto v : e[u]) {
    if (v == son[u]) continue;
    int sizv = vec[v]->size();
    for (int i = 0; i < sizv; ++i) {
      (*vec[u])[sizu - sizv + i] += (*vec[v])[i];
    }
  }
  vec[u]->push_back(1);
  sizu += 1;
  if (u == 0) return;
  for (auto &p : query[u]) {
    if (p.first >= sizu) {
      p.second = 0;
    } else {
      p.second = (*vec[u])[sizu - p.first - 1] - 1;
    }
  }
}
int askv[100015], askp[100015];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int p;
    scanf("%d", &p);
    e[p].push_back(i);
  }
  dfs1(0);
  pre();
  int q;
  scanf("%d", &q);
  for (int i = 1; i <= q; ++i) {
    scanf("%d%d", &askv[i], &askp[i]);
    askv[i] = query_pth_anc(askv[i], askp[i]);
    if (~askv[i]) query[askv[i]][askp[i]];
  }
  dfs2(0);
  for (int i = 1; i <= q; ++i) {
    printf("%lld ", ~askv[i] ? query[askv[i]][askp[i]] : 0);
  }
  return 0;
}