Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long int N = 1e5 + 10, N2 = 1e5, delta = 10501, mod = 1e9 + 7,
                    oo = 1e9 + 10, LOG = 20, SQ = 300;
const long double PI = 3.141592653589793;
long long int a[N], par[LOG + 2][N], h[N], ans[N], siz[N];
vector<long long int> v[N];
map<long long int, long long int> mp[N];
vector<pair<long long int, long long int> > q[N];
void dfspar(long long int x) {
  for (long long int i = 1; i <= LOG; i++)
    par[i][x] = par[i - 1][par[i - 1][x]];
  for (long long int u : v[x]) {
    h[u] = h[x] + 1;
    par[0][u] = x;
    dfspar(u);
  }
}
long long int findpar(long long int u, long long int x) {
  for (long long int i = LOG; ~i; i--)
    if (x >> i & 1) u = par[i][u];
  return u;
}
void dfs(long long int x) {
  siz[x] = mp[x][h[x]] = 1;
  for (long long int u : v[x]) {
    dfs(u);
    if (siz[u] > siz[x]) {
      mp[u].swap(mp[x]);
      swap(siz[u], siz[x]);
    }
    for (auto u2 : mp[u]) mp[x][u2.first] += mp[u][u2.first];
    siz[x] += siz[u];
  }
  for (auto u : q[x]) ans[u.first] = mp[x][u.second];
}
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long int n;
  cin >> n;
  for (long long int i = 1; i <= n; i++) {
    cin >> a[i];
    v[a[i]].push_back(i);
  }
  for (long long int i = 1; i <= n; i++)
    if (!a[i]) dfspar(i);
  long long int m;
  cin >> m;
  for (long long int i = 0; i < m; i++) {
    long long int v, p;
    cin >> v >> p;
    q[findpar(v, p)].push_back({i, h[v]});
  }
  for (long long int i = 1; i <= n; i++)
    if (!a[i]) dfs(i);
  for (long long int i = 0; i < m; i++) cout << max(1ll, ans[i]) - 1 << ' ';
  return 0;
}