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;
int n;
struct hh {
  int t, nxt;
} edge[1000020 << 1];
int head[1000020], ecnt;
void make_edge(int f, int t) {
  edge[++ecnt] = (hh){t, head[f]};
  head[f] = ecnt;
  edge[++ecnt] = (hh){f, head[t]};
  head[t] = ecnt;
}
int dep[1000020], size[1000020], son[1000020], fa[1000020];
void dfs1(int x, int fa) {
  dep[x] = dep[fa] + 1;
  size[x] = 1;
  for (int i = head[x]; i; i = edge[i].nxt)
    if (edge[i].t != fa) {
      dfs1(edge[i].t, x);
      size[x] += size[edge[i].t];
      if (size[edge[i].t] > size[son[x]]) son[x] = edge[i].t;
    }
}
map<string, int> S[1000020];
int sum[1000020];
string a[1000020];
bool s[1000020];
void add(string s, int d, int t) {
  if (!S[d][s]) ++sum[d];
  S[d][s] += t;
  if (!S[d][s]) --sum[d];
}
void add(int x, int fa, int t) {
  add(a[x], dep[x], t);
  for (int i = head[x]; i; i = edge[i].nxt)
    if (edge[i].t != fa && !s[edge[i].t]) add(edge[i].t, x, t);
}
vector<pair<int, int> > q[1000020];
int ans[1000020];
void dfs2(int x, int fa, bool keep) {
  for (int i = head[x]; i; i = edge[i].nxt)
    if (edge[i].t != fa && edge[i].t != son[x]) dfs2(edge[i].t, x, 0);
  s[son[x]] = 1;
  if (son[x]) dfs2(son[x], x, 1);
  add(x, fa, 1);
  for (int i = 0; i < (int)q[x].size(); i++)
    ans[q[x][i].first] = sum[q[x][i].second];
  s[son[x]] = 0;
  if (!keep) add(x, fa, -1);
}
int main() {
  ios::sync_with_stdio(false);
  int i, x, y, m;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> a[i] >> fa[i];
    if (fa[i]) make_edge(fa[i], i);
  }
  for (i = 1; i <= n; i++)
    if (!fa[i]) dfs1(i, 0);
  cin >> m;
  for (i = 1; i <= m; i++) {
    cin >> x >> y;
    q[x].push_back(make_pair(i, dep[x] + y));
  }
  for (i = 1; i <= n; i++)
    if (!fa[i]) dfs2(i, 0, 0);
  for (i = 1; i <= m; i++) printf("%d\n", ans[i]);
}
