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 + 5;
const int mod = 1e9 + 7;
long long qpow(long long a, long long b) {
  long long res = 1;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
struct graph {
  int head[maxn], nxt[maxn << 1], to[maxn << 1], w[maxn << 1], sz;
  void init() { memset(head, -1, sizeof(head)); }
  graph() { init(); }
  void push(int a, int b, int c = 0) {
    nxt[sz] = head[a], to[sz] = b, w[sz] = c, head[a] = sz++;
  }
  int& operator[](const int a) { return to[a]; }
} g;
int pre[20][maxn], dep[maxn], dfn[maxn], l[maxn], r[maxn], num, rnk[maxn];
int n;
void dfs(int now, int fa) {
  l[now] = ++num;
  dfn[now] = num;
  rnk[num] = now;
  dep[now] = dep[fa] + 1;
  pre[0][now] = fa;
  for (int i = 1; i < 20; ++i) {
    pre[i][now] = pre[i - 1][pre[i - 1][now]];
  }
  for (int i = g.head[now]; ~i; i = g.nxt[i]) {
    dfs(g[i], now);
  }
  r[now] = num;
}
struct node {
  int l, r, cnt;
};
int root[maxn];
struct Segment_tree {
  node tree[maxn * 40];
  int tot;
  void Insert(int& rt, int pre, int l, int r, int id) {
    rt = ++tot;
    tree[rt] = tree[pre];
    tree[rt].cnt++;
    if (l == r) return;
    int mid = l + r >> 1;
    if (mid >= id)
      Insert(tree[rt].l, tree[pre].l, l, mid, id);
    else
      Insert(tree[rt].r, tree[pre].r, mid + 1, r, id);
  }
  int query(int rt, int l, int r, int id) {
    if (l == r) {
      return tree[rt].cnt;
    }
    int mid = l + r >> 1;
    if (mid >= id) return query(tree[rt].l, l, mid, id);
    return query(tree[rt].r, mid + 1, r, id);
  }
} sg;
int cal(int x, int y) {
  if (dep[x] - y <= 0) return 0;
  for (int i = 18; i >= 0; --i) {
    if (y >> i & 1) {
      x = pre[i][x];
    }
  }
  return x;
}
int main() {
  scanf("%d", &n);
  for (int i = 1, f; i <= n; ++i) {
    scanf("%d", &f);
    g.push(f, i);
  }
  dep[0] = -1;
  dfs(0, 0);
  for (int i = 1; i <= num; ++i) {
    sg.Insert(root[i], root[i - 1], 0, n, dep[rnk[i]]);
  }
  int q;
  scanf("%d", &q);
  int v, p;
  while (q--) {
    scanf("%d%d", &v, &p);
    int fa = cal(v, p);
    if (fa == 0) {
      printf("0 ");
      continue;
    }
    int ql = l[fa], qr = r[fa];
    printf("%d ", sg.query(root[qr], 0, n, dep[v]) -
                      sg.query(root[ql - 1], 0, n, dep[v]) - 1);
  }
  return 0;
}
