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 N = 1e5 + 5;
vector<int> adj[N];
int c[N], h[N], sz[N], isbig[N], vis[N], PR[N][19];
int mp[N];
vector<pair<int, int> > query[N];
void countsize(int v, int p) {
  sz[v] = 1;
  h[v] = h[p] + 1;
  PR[v][0] = p;
  for (int h = 1; h < 19; h++) PR[v][h] = PR[PR[v][h - 1]][h - 1];
  for (int i : adj[v])
    if (i != p) {
      countsize(i, v);
      sz[v] += sz[i];
    }
}
int ans[N], A[N];
void add(int v, int p, int val) {
  mp[h[v]] += val;
  for (int i : adj[v])
    if (!isbig[i] && i != p) add(i, v, val);
}
void dfs(int v, int p, bool keep) {
  int big = -1, mxsz = -1;
  for (int i : adj[v])
    if (i != p) {
      if (sz[i] > mxsz) {
        mxsz = sz[i];
        big = i;
      }
    }
  for (int i : adj[v])
    if (i != p && i != big) dfs(i, v, 0);
  if (big + 1) {
    dfs(big, v, 1);
    isbig[big] = 1;
  }
  add(v, p, 1);
  isbig[big] = 0;
  for (auto i : query[v]) {
    int H = i.first;
    int q = i.second;
    A[q] = (H < N ? mp[H] : 1);
  }
  if (keep == 0) add(v, p, -1);
}
int P[N];
int pr(int v, int k) {
  for (int i = 0; i < 19; i++)
    if ((1 << i) & k) v = PR[v][i];
  return v;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int p;
    cin >> p;
    P[i] = p;
    adj[p].push_back(i);
  }
  cin >> m;
  for (int i = 1; i <= n; i++)
    if (P[i] == 0) countsize(i, 0);
  for (int i = 1; i <= m; i++) {
    A[i] = 1;
    int v, k;
    cin >> v >> k;
    if (h[v] <= k)
      A[i] = 1;
    else
      query[pr(v, k)].push_back({h[v], i});
  }
  for (int i = 1; i <= n; i++)
    if (P[i] == 0) dfs(i, 0, 0);
  for (int i = 1; i <= m; i++) cout << A[i] - 1 << ' ';
}
