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;
int n, m, tot, pre[200010 << 1], ans[200010], now[200010], d[200010],
    son[200010 << 1], z[200010], s[200010], g[200010];
map<string, int> c[200010];
string ch[200010];
vector<int> a[200010], b[200010];
void put(int x, int y) {
  pre[++tot] = now[x];
  now[x] = tot;
  son[tot] = y;
}
void dfs1(int u, int fa) {
  d[u] = d[fa] + 1, s[u] = 1;
  int num = 0;
  for (int i = now[u]; i; i = pre[i])
    if (son[i] != fa) {
      dfs1(son[i], u), s[u] += s[son[i]];
      if (num < s[son[i]]) num = s[son[i]], z[u] = son[i];
    }
}
void qk(int u, int fa) {
  if (u != 0) {
    c[d[u]][ch[u]]--;
    if (!c[d[u]][ch[u]]) g[d[u]]--;
  }
  for (int i = now[u]; i; i = pre[i])
    if (son[i] != fa) qk(son[i], u);
}
void dfs3(int u, int fa, int zz) {
  if (u != 0) {
    if (!c[d[u]][ch[u]]) g[d[u]]++;
    c[d[u]][ch[u]]++;
  }
  for (int i = now[u]; i; i = pre[i])
    if (son[i] != fa && son[i] != zz) dfs3(son[i], u, zz);
}
void dfs2(int u, int fa) {
  for (int i = now[u]; i; i = pre[i])
    if (son[i] != fa && son[i] != z[u]) dfs2(son[i], u), qk(son[i], u);
  if (z[u]) dfs2(z[u], u);
  dfs3(u, fa, z[u]);
  for (int i = 0; i < a[u].size(); i++) ans[b[u][i]] = g[a[u][i] + d[u]];
}
signed main() {
  scanf("%d", &n);
  for (int i = 1, x; i <= n; i++)
    cin >> ch[i], scanf("%d", &x), put(x, i), put(i, x);
  scanf("%d", &m);
  for (int i = 1, x, y; i <= m; i++)
    scanf("%d%d", &x, &y), a[x].push_back(y), b[x].push_back(i);
  dfs1(0, 0);
  dfs2(0, 0);
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}
