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 hh {
  int p, w;
};
map<string, int> q;
int a[100005];
vector<hh> qr[100005];
vector<int> d[100005];
int dis[100005], l[100005], r[100005], dfn, sz[100005], son[100005], id[100005],
    mx, ans[100005];
set<int> sum[100005];
void calc(int x, int f) {
  sz[x] = 1;
  l[x] = ++dfn;
  id[dfn] = x;
  dis[x] = dis[f] + 1;
  mx = max(mx, dis[x]);
  for (int i = 0; i < (int)d[x].size(); ++i) {
    int v = d[x][i];
    calc(v, x);
    sz[x] += sz[v];
    if (sz[v] > sz[son[x]]) son[x] = v;
  }
  r[x] = dfn;
}
void xg(int x) { sum[dis[x]].insert(a[x]); }
void dfs(int x) {
  for (int i = 0; i < (int)d[x].size(); ++i) {
    int v = d[x][i];
    if (v == son[x]) continue;
    dfs(v);
    for (int j = l[v]; j <= r[v]; ++j) sum[dis[id[j]]].clear();
  }
  if (son[x]) dfs(son[x]);
  for (int i = 0; i < (int)d[x].size(); ++i) {
    int v = d[x][i];
    if (v == son[x]) continue;
    for (int j = l[v]; j <= r[v]; ++j) xg(id[j]);
  }
  xg(x);
  for (int i = 0; i < (int)qr[x].size(); ++i)
    ans[qr[x][i].p] = sum[dis[x] + qr[x][i].w].size();
}
int main() {
  int n, m, cnt;
  string b;
  scanf("%d", &n);
  for (int i = 1, x; i <= n; ++i) {
    cin >> b >> x;
    if (!q[b]) q[b] = ++cnt;
    a[i] = q[b];
    d[x].push_back(i);
  }
  scanf("%d", &m);
  for (int i = 1, a, b; i <= m; ++i) {
    scanf("%d%d", &a, &b);
    qr[a].push_back((hh){i, b});
  }
  calc(0, 0);
  dfs(0);
  for (int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
}
