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;
bool root[100010];
vector<int> child[100010];
string name[100010];
map<string, int> mp;
int nameid[100010];
char buf[30];
int X;
int L[100010], R[100010], depth[100010];
vector<int> level[100010];
int Q;
int qroot[100010], qdepth[100010];
vector<pair<pair<int, int>, int> > queries;
int ans[100010];
int dfs(int x, int d) {
  int i;
  depth[x] = d;
  level[d].push_back(x);
  L[x] = X;
  X++;
  for ((i) = 0; (i) < (int)(child[x].size()); (i)++) dfs(child[x][i], d + 1);
  R[x] = X;
  X++;
}
int count_smaller(int depth, int bound) {
  int sz = level[depth].size();
  int low = 0, high = sz + 1;
  while (high - low > 1) {
    int mid = (high + low) / 2;
    if (L[level[depth][mid - 1]] < bound)
      low = mid;
    else
      high = mid;
  }
  return low;
}
map<pair<int, int>, int> cache;
bool used[100010];
int query(int root, int depth) {
  if (depth > 100005) return 0;
  pair<int, int> state = make_pair(root, depth);
  if (cache.find(state) != cache.end()) return cache[state];
  int low = count_smaller(depth, L[root]);
  int high = count_smaller(depth, R[root]);
  if (low >= high) return 0;
  int ans = 0;
  int i;
  for (i = low; i < high; i++) {
    int x = nameid[level[depth][i]];
    if (!used[x]) {
      ans++;
      used[x] = true;
    }
  }
  for (i = low; i < high; i++) {
    int x = nameid[level[depth][i]];
    used[x] = false;
  }
  return cache[state] = ans;
}
void solve(void) {
  int i;
  for ((i) = 0; (i) < (int)(N); (i)++)
    if (root[i]) dfs(i, 0);
  for ((i) = 0; (i) < (int)(Q); (i)++)
    ans[i] = query(qroot[i], qdepth[i] + depth[qroot[i]]);
}
int main(void) {
  int i;
  scanf("%d", &N);
  for ((i) = 0; (i) < (int)(N); (i)++) {
    scanf("%s", buf);
    name[i] = buf;
    int p;
    scanf("%d", &p);
    if (p == 0)
      root[i] = true;
    else
      child[p - 1].push_back(i);
  }
  scanf("%d", &Q);
  for ((i) = 0; (i) < (int)(Q); (i)++) {
    scanf("%d%d", &qroot[i], &qdepth[i]);
    qroot[i]--;
  }
  for ((i) = 0; (i) < (int)(N); (i)++) mp[name[i]] = 0;
  i = 0;
  for (__typeof((mp).begin()) itr = (mp).begin(); itr != (mp).end(); itr++) {
    (itr->second) = i;
    i++;
  }
  for ((i) = 0; (i) < (int)(N); (i)++) nameid[i] = mp[name[i]];
  solve();
  for ((i) = 0; (i) < (int)(Q); (i)++) printf("%d\n", ans[i]);
  return 0;
}
