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 INF = 0x3f3f3f3f;
const int MAXN = 1e5 + 5;
const int LOG = 20;
int n, m, len, tot;
int head[MAXN], root[MAXN], vis[MAXN], son[MAXN], size[MAXN], dep[MAXN],
    f[MAXN][LOG], ans[MAXN], cnt[MAXN];
struct Edge {
  int to, next;
} e[MAXN << 1];
struct Query {
  int x, id;
};
vector<Query> vec[MAXN];
void add(int u, int v) {
  e[++len].to = v;
  e[len].next = head[u];
  head[u] = len;
}
void dfs(int u, int fa, int d) {
  dep[u] = d;
  f[u][0] = fa;
  for (int i = 1; i < LOG; i++) f[u][i] = f[f[u][i - 1]][i - 1];
  size[u] = 1;
  for (int i = head[u]; i != -1; i = e[i].next) {
    int v = e[i].to;
    if (v == fa) continue;
    dfs(v, u, d + 1);
    size[u] += size[v];
    if (size[v] > size[son[u]]) son[u] = v;
  }
}
void Calc(int u, int fa, int val) {
  cnt[dep[u]] += val;
  for (int i = head[u]; i != -1; i = e[i].next) {
    int v = e[i].to;
    if (v == fa || vis[v]) continue;
    Calc(v, u, val);
  }
}
void Dsu(int u, int fa, int flag) {
  for (int i = head[u]; i != -1; i = e[i].next) {
    int v = e[i].to;
    if (v == fa || v == son[u]) continue;
    Dsu(v, u, 0);
  }
  if (son[u]) {
    Dsu(son[u], u, 1);
    vis[son[u]] = 1;
  }
  Calc(u, fa, 1);
  for (int i = 0; i < (int)vec[u].size(); i++)
    ans[vec[u][i].id] = cnt[vec[u][i].x] - 1;
  vis[son[u]] = 0;
  if (!flag) Calc(u, fa, -1);
}
int main() {
  memset(head, -1, sizeof(head));
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    if (x)
      add(x, i);
    else
      root[++tot] = i;
  }
  for (int i = 1; i <= tot; i++) dfs(root[i], 0, 1);
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    int x, k;
    scanf("%d%d", &x, &k);
    int d = dep[x];
    for (int i = LOG - 1; i >= 0; i--) {
      if (k & (1 << i)) x = f[x][i];
    }
    vec[x].push_back((Query){d, i});
  }
  for (int i = 1; i <= tot; i++) Dsu(root[i], 0, 0);
  for (int i = 1; i <= m; i++) printf("%d ", ans[i]);
  return 0;
}
