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 = 100000;
int n, q;
string st[N + 10];
vector<int> xw[N + 10], xb[N + 10];
int son[N * 2 + 10], head[N + 10], nxt[N * 2 + 10], tot, f[N + 10];
void add(int x, int y) { son[++tot] = y, nxt[tot] = head[x], head[x] = tot; }
int sz[N + 10], dep[N + 10];
int ans[N + 10];
int bs[N + 10], bigson;
int len[N + 10];
map<string, bool> t[N + 10];
bool vis[N + 10];
void tadd(int x, int p) {
  if (p == 1 && !t[dep[x]][st[x]]) t[dep[x]][st[x]] = 1, len[dep[x]]++;
  if (p == -1) t[dep[x]].clear(), len[dep[x]] = 0;
  for (int i = head[x]; i; i = nxt[i])
    if (son[i] != bigson) tadd(son[i], p);
}
void dfs(int x, bool ke) {
  vis[x] = 1;
  for (int i = head[x]; i; i = nxt[i])
    if (son[i] != bs[x]) dfs(son[i], 0);
  if (bs[x]) dfs(bs[x], 1), bigson = bs[x];
  tadd(x, 1);
  bigson = 0;
  int len_ = xb[x].size();
  for (int i = 0; i < len_; i++) ans[xb[x][i]] = len[xw[x][i] + dep[x]];
  if (ke == 0) tadd(x, -1);
}
void dfs1(int x, int fa) {
  sz[x] = 1, dep[x] = dep[fa] + 1;
  int mx = -1;
  for (int i = head[x]; i; i = nxt[i]) {
    dfs1(son[i], x);
    sz[x] += sz[son[i]];
    if (sz[son[i]] > mx) {
      mx = sz[son[i]];
      bs[x] = son[i];
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1, x; i <= n; i++) {
    cin >> st[i];
    scanf("%d", &f[i]);
    if (f[i] != 0) add(f[i], i);
  }
  for (int i = 1; i <= n; i++)
    if (f[i] == 0) dfs1(i, 0);
  scanf("%d", &q);
  for (int i = 1, x, y; i <= q; i++) {
    scanf("%d %d", &x, &y);
    if (dep[x] + y <= n) xw[x].push_back(y), xb[x].push_back(i);
  }
  for (int i = 1; i <= n; i++)
    if (f[i] == 0) dfs(i, 0);
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
  return 0;
}
