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;
struct Query {
  int d, id;
};
const int MAXN = 1e5 + 50;
const int LOGN = 17;
vector<int> adj[MAXN];
int pa[MAXN][LOGN], ans[MAXN], n, q;
vector<Query> qlist[MAXN];
void readInput() {
  scanf("%d", &n);
  for (int v = 1; v <= n; v++) {
    scanf("%d", &pa[v][0]);
    adj[pa[v][0]].push_back(v);
  }
  pa[0][0] = 0;
  for (int i = 1; i < LOGN; i++)
    for (int v = 0; v <= n; v++) pa[v][i] = pa[pa[v][i - 1]][i - 1];
}
int sz[MAXN], depth[MAXN], tin[MAXN], tout[MAXN], vin[2 * MAXN];
int CUR_TIME = 0;
void preprocess(int v) {
  tin[v] = ++CUR_TIME;
  vin[tin[v]] = v;
  sz[v] = 1;
  for (auto c : adj[v]) {
    depth[c] = depth[v] + 1;
    preprocess(c);
    sz[v] += sz[c];
  }
  tout[v] = ++CUR_TIME;
}
void readQuery() {
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int v, d;
    scanf("%d%d", &v, &d);
    if (depth[v] <= d)
      ans[i] = 0;
    else {
      for (int x = 0; x < LOGN; x++)
        if (d & (1 << x)) v = pa[v][x];
      qlist[v].push_back({d, i});
    }
  }
}
int cnt[MAXN];
void dfs(int v, bool keep) {
  int big = -1;
  for (auto c : adj[v])
    if (big == -1 || sz[c] > sz[big]) big = c;
  for (auto c : adj[v])
    if (c != big) dfs(c, false);
  if (big != -1) dfs(big, true);
  for (auto c : adj[v])
    if (c != big) {
      for (int t = tin[c]; t < tout[c]; t++)
        if (vin[t]) cnt[depth[vin[t]]]++;
    }
  cnt[depth[v]]++;
  for (auto q : qlist[v]) ans[q.id] = cnt[depth[v] + q.d] - 1;
  if (!keep) {
    for (int t = tin[v]; t < tout[v]; t++)
      if (vin[t]) cnt[depth[vin[t]]]--;
  }
}
int main() {
  readInput();
  preprocess(0);
  readQuery();
  dfs(0, false);
  for (int i = 1; i <= q; i++) printf("%d ", ans[i]);
}
