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 data {
  int id, x;
};
int n, m, now, o1, o2, mp0;
int a[(100010)], f[(100010)], son[(100010)], sz[(100010)], id[(100010)],
    l[(100010)], r[(100010)], dep[(100010)];
int cnt[(100010)], res[(100010)], ans[(100010)];
string o;
vector<int> edge[(100010)];
vector<data> q[(100010)];
map<string, int> mp;
map<int, int> mp1[(100010)];
void dfs1(int x) {
  sz[x] = 1;
  l[x] = ++now;
  id[now] = x;
  dep[x] = dep[f[x]] + 1;
  for (int i = 0; i < edge[x].size(); i++) {
    int y = edge[x][i];
    dfs1(edge[x][i]);
    sz[x] += sz[y];
    if (sz[son[x]] < sz[y]) son[x] = y;
  }
  r[x] = now;
}
void del(int x) {
  for (int i = l[x]; i <= r[x]; i++) {
    --mp1[dep[id[i]]][a[id[i]]];
    if (!mp1[dep[id[i]]][a[id[i]]]) --res[dep[id[i]]];
  }
}
void add(int x) {
  for (int i = l[x]; i <= r[x]; i++) {
    if (!mp1[dep[id[i]]][a[id[i]]]) ++res[dep[id[i]]];
    ++mp1[dep[id[i]]][a[id[i]]];
  }
}
void dfs2(int x) {
  for (int i = 0; i < edge[x].size(); i++) {
    int y = edge[x][i];
    if (y != son[x]) dfs2(y), del(y);
  }
  if (son[x]) dfs2(son[x]);
  for (int i = 0; i < edge[x].size(); i++) {
    int y = edge[x][i];
    if (y != son[x]) add(y);
  }
  if (!mp1[dep[x]][a[x]]) ++res[dep[x]];
  ++mp1[dep[x]][a[x]];
  for (int i = 0; i < q[x].size(); i++) {
    ans[q[x][i].id] = res[q[x][i].x];
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n + 1; i++) {
    cin >> o;
    scanf("%d", &o1);
    ++o1;
    if (!mp[o]) mp[o] = ++mp0;
    a[i] = mp[o];
    edge[o1].push_back(i);
    f[i] = o1;
  }
  dfs1(1);
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &o1, &o2);
    q[o1 + 1].push_back({i, o2 + dep[o1 + 1]});
  }
  dfs2(1);
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
}
