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 Maxn = 200005;
int n, q, dfs_time;
int st[Maxn], ed[Maxn], to[Maxn << 2];
int dep[Maxn], ans[Maxn];
int qry[Maxn][2];
vector<int> g[Maxn];
vector<pair<int, int> > qp[Maxn << 2];
vector<int> que[Maxn << 2];
void dfs(int u, int depth) {
  st[u] = ++dfs_time;
  dep[u] = depth;
  to[dfs_time] = u;
  for (auto &v : g[u]) dfs(v, depth + 1);
  ed[u] = ++dfs_time;
  to[dfs_time] = u;
}
int stk[Maxn];
void dfs2(int u, int h) {
  stk[h] = u;
  for (auto &id : que[u]) {
    int p = qry[id][1];
    int lca = stk[h - p];
    qp[st[lca] - 1].push_back({id, -1});
    qp[ed[lca]].push_back({id, 1});
  }
  for (auto &v : g[u]) dfs2(v, h + 1);
}
int cnt[Maxn << 2];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int k;
    scanf("%d", &k);
    g[k].push_back(i);
  }
  dfs(0, 0);
  scanf("%d", &q);
  for (int i = 0; i < q; ++i) {
    scanf("%d %d", qry[i], qry[i] + 1);
    que[*qry[i]].push_back(i);
  }
  dfs2(0, 0);
  for (int i = 1; i < dfs_time; ++i) {
    cnt[dep[to[i]]]++;
    for (auto [id, f] : qp[i]) ans[id] += cnt[dep[qry[id][0]]] * f;
  }
  for (int i = 0; i < q; ++i)
    printf("%d%c", max(ans[i] / 2 - 1, 0), " \n"[i == q - 1]);
  return 0;
}
