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 = 1e5 + 10;
int n;
int fa[MAXN], son[MAXN], dep[MAXN], siz[MAXN];
int col[MAXN];
int cnt, head[MAXN];
struct Edge {
  int to, next;
} e[MAXN << 1];
struct Ex {
  int x, c;
};
vector<Ex> ex[MAXN];
void addedge(int u, int v) {
  e[cnt].to = v;
  e[cnt].next = head[u];
  head[u] = cnt++;
  e[cnt].to = u;
  e[cnt].next = head[v];
  head[v] = cnt++;
}
void dfs(int u) {
  siz[u] = 1;
  son[u] = 0;
  for (int i = head[u]; ~i; i = e[i].next) {
    if (e[i].to != fa[u]) {
      fa[e[i].to] = u;
      dep[e[i].to] = dep[u] + 1;
      dfs(e[i].to);
      if (siz[e[i].to] > siz[son[u]]) son[u] = e[i].to;
      siz[u] += siz[e[i].to];
    }
  }
}
int vis[MAXN], ans[MAXN];
char name[MAXN][22];
set<string> S[MAXN << 1];
void change(int u, int c) {
  if (c > 0) {
    S[dep[u]].insert(name[u]);
  } else
    S[dep[u]].clear();
  for (int i = head[u]; ~i; i = e[i].next) {
    if (e[i].to != fa[u] && !vis[e[i].to]) change(e[i].to, c);
  }
}
void dfs1(int u, int flg) {
  for (int i = head[u]; ~i; i = e[i].next) {
    if (e[i].to != fa[u] && e[i].to != son[u]) dfs1(e[i].to, 1);
  }
  if (son[u]) {
    dfs1(son[u], 0);
    vis[son[u]] = 1;
  }
  change(u, 1);
  int sz = ex[u].size();
  for (int i = 0; i < sz; i++) {
    ans[ex[u][i].x] = S[ex[u][i].c].size();
  }
  if (son[u]) vis[son[u]] = 0;
  if (flg) change(u, -1);
}
int main() {
  scanf("%d", &n);
  memset(head, -1, sizeof head);
  cnt = 0;
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%s%d", name[i], &x);
    addedge(i, x);
  }
  dfs(0);
  int m;
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    ex[x].push_back(Ex{i, dep[x] + y});
  }
  dfs1(0, 0);
  for (int i = 0; i < m; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}
