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 inf = 0x3f3f3f3f;
const int maxn = 2e5 + 5;
const long long mod = 998244353;
unordered_map<int, unordered_map<string, int>> tmp;
int t, n, m, k1, k, tim, num, sum;
int dui[maxn], dfn[maxn], siz[maxn], vis[maxn], id[maxn], ans[maxn], a[maxn],
    son[maxn], dep[maxn];
char h[100005][25];
struct node {
  int id, dep;
};
vector<node> que[maxn];
vector<int> mp[maxn];
void dfs1(int x) {
  siz[x] = 1;
  dfn[x] = ++tim;
  id[tim] = x;
  for (int i = 0; i < mp[x].size(); i++) {
    int y = mp[x][i];
    dep[y] = dep[x] + 1;
    dfs1(y);
    siz[x] += siz[y];
    if (siz[y] > siz[son[x]]) son[x] = y;
  }
}
void dfs(int x, int op) {
  for (int i = 0; i < mp[x].size(); i++) {
    int y = mp[x][i];
    if (y != son[x]) dfs(y, 0);
  }
  if (son[x]) dfs(son[x], 1);
  for (int i = 0; i < mp[x].size(); i++) {
    int v = mp[x][i];
    if (v == son[x]) continue;
    for (int j = dfn[v]; j <= dfn[v] + siz[v] - 1; j++) {
      int y = id[j];
      string p(h[y], h[y] + strlen(h[y]));
      tmp[dep[y]][p]++;
    }
  }
  string p(h[x], h[x] + strlen(h[x]));
  tmp[dep[x]][p]++;
  for (int i = 0; i < que[x].size(); i++) {
    ans[que[x][i].id] = tmp[que[x][i].dep].size();
  }
  if (!op) tmp.clear();
}
int main() {
  scanf("%d", &n);
  for (int i = 1, x; i <= n; i++) {
    scanf("%s%d", h[i], &x);
    if (x == 0)
      dui[++t] = i;
    else
      mp[x].push_back(i);
  }
  for (int i = 1; i <= t; i++) dfs1(dui[i]);
  scanf("%d", &m);
  for (int i = 1, x, y; i <= m; i++) {
    scanf("%d%d", &x, &y);
    que[x].push_back({i, dep[x] + y});
  }
  for (int i = 1; i <= t; i++) dfs(dui[i], 0);
  for (int i = 1; i <= m; i++) {
    printf("%d\n", ans[i]);
  }
}
