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 N = 1e5 + 5;
const int LogN = 17 + 1;
vector<int> g[N];
int pr[LogN][N];
int tin[N];
int tout[N];
int dist[N];
int n, q;
void dfs(int u) {
  static int timer = 0;
  tin[u] = ++timer;
  for (int v : g[u]) {
    if (!tin[v]) {
      dist[v] = dist[u] + 1;
      dfs(v);
    }
  }
  tout[u] = timer;
}
struct segment_tree {
  struct node {
    map<int, int> cnt;
    node() {}
  };
  node st[N << 2];
  void update(int x, int l, int r, int k, int v) {
    st[x].cnt[v]++;
    if (l == r) {
      return;
    }
    int mid = (l + r) / 2;
    if (k <= mid) {
      update(x << 1, l, mid, k, v);
    } else {
      update(x << 1 | 1, mid + 1, r, k, v);
    }
  }
  void update(int u) { update(1, tin[0], tout[0], tin[u], dist[u]); }
  int query(int x, int l, int r, int ll, int rr, int v) {
    if (r < ll || rr < l || r < l) {
      return 0;
    }
    if (ll <= l && r <= rr) {
      return st[x].cnt.count(v) ? st[x].cnt[v] : 0;
    }
    int mid = (l + r) / 2;
    return query(x << 1, l, mid, ll, rr, v) +
           query(x << 1 | 1, mid + 1, r, ll, rr, v);
  }
  int query(int z, int u) {
    return query(1, tin[0], tout[0], tin[z], tout[z], dist[u]);
  }
};
segment_tree st;
void build_lca() {
  for (int k = (1); k <= ((LogN - 1)); k += (+1)) {
    for (int u = (1); u <= ((n)); u += (+1)) {
      if (pr[k - 1][u] != 0) {
        pr[k][u] = pr[k - 1][pr[k - 1][u]];
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0), cout.precision(15);
  cin >> n;
  for (int u = (1); u <= ((n)); u += (+1)) {
    cin >> pr[0][u];
    g[pr[0][u]].push_back(u);
  }
  dfs(0);
  build_lca();
  for (int u = (1); u <= ((n)); u += (+1)) {
    st.update(u);
  }
  cin >> q;
  while (q--) {
    int u, p_anc;
    cin >> u >> p_anc;
    int cousins = 0;
    if (dist[u] > p_anc) {
      int z = u;
      for (int k = (LogN - 1); k >= (0); k += (-1)) {
        if ((1 << k) <= p_anc) {
          p_anc -= (1 << k);
          z = pr[k][z];
        }
      }
      cousins = st.query(z, u) - 1;
    }
    cout << cousins << " \n"[q == 0];
  }
  return 0;
}
