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 int N = 200100;
int n, m, sz[N], deep[N], ans[N], mod = 9576890767, hsh[N];
char A[30];
string second;
vector<int> adj[N], root;
vector<pair<int, int> > query[N];
set<int> s[N];
void calcSz(int u, int v) {
  sz[u] = 1;
  deep[u] = deep[v] + 1;
  for (auto &x : adj[u]) {
    if (x != v) {
      calcSz(x, u);
      sz[u] += sz[x];
    }
  }
}
void add(int u, int p) {
  if (p == 1)
    s[deep[u]].insert(hsh[u]);
  else
    s[deep[u]].erase(hsh[u]);
}
void dfs2(int u, int v, int p) {
  add(u, p);
  for (auto &x : adj[u])
    if (x != v) dfs2(x, u, p);
}
void dfs(int u, int v, bool keep) {
  int mxSz = 0, bigChild = 0;
  for (auto &x : adj[u])
    if (x != v && mxSz < sz[x]) {
      mxSz = sz[x];
      bigChild = x;
    }
  for (auto &x : adj[u])
    if (x != v && x != bigChild) dfs(x, u, 0);
  if (bigChild != 0) dfs(bigChild, u, 1);
  for (auto &x : adj[u])
    if (x != v && x != bigChild) dfs2(x, u, 1);
  add(u, 1);
  for (auto &X : query[u]) {
    int x = X.first;
    int id = X.second;
    ans[id] = (int)(s[deep[u] + x].size());
  }
  if (!keep) dfs2(u, v, -1);
}
int getHash(string &s, int mod) {
  int ret = 1;
  for (auto &x : s) ret = (1LL * ret * 26 + (x - 'a')) % mod;
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%s", &A);
    second = A;
    hsh[i] = getHash(second, mod);
    int x;
    scanf("%d", &x);
    if (x) {
      adj[i].push_back(x);
      adj[x].push_back(i);
    } else
      root.push_back(i);
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    int u, h;
    scanf("%d%d", &u, &h);
    query[u].push_back({h, i});
  }
  for (auto &x : root) {
    deep[x] = -1;
    calcSz(x, x);
    dfs(x, x, 0);
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
}