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 inf = 0x3fffffff;
const long long mod = 1000000007;
const int maxn = 1e5 + 100;
int head[maxn];
string s[maxn];
vector<pair<int, int> > q[maxn];
struct edge {
  int to, next;
} e[maxn];
int tol = 0;
void add(int u, int v) {
  e[++tol].to = v, e[tol].next = head[u], head[u] = tol;
}
int sz[maxn], son[maxn], deep[maxn], mark[maxn];
int cnt[maxn];
set<string> st[maxn * 2];
int ans[maxn];
void dfs1(int u) {
  sz[u] = 1, son[u] = 0;
  for (int i = head[u]; i; i = e[i].next) {
    int v = e[i].to;
    deep[v] = deep[u] + 1;
    dfs1(v);
    sz[u] += sz[v];
    if (!son[u] || sz[v] > sz[son[u]]) son[u] = v;
  }
}
void cal(int u, int val) {
  if (val > 0)
    st[deep[u]].insert(s[u]);
  else
    st[deep[u]].clear();
  for (int i = head[u]; i; i = e[i].next) {
    int v = e[i].to;
    if (!mark[v]) cal(v, val);
  }
}
void dfs(int u, int keep) {
  for (int i = head[u]; i; i = e[i].next) {
    int v = e[i].to;
    if (v != son[u]) dfs(v, 0);
  }
  if (son[u]) dfs(son[u], 1), mark[son[u]] = 1;
  cal(u, 1);
  for (int i = 0; i < q[u].size(); i++) {
    pair<int, int> t = q[u][i];
    ans[t.second] = (int)st[deep[u] + t.first].size();
  }
  if (son[u]) mark[son[u]] = 0;
  if (!keep) cal(u, -1);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n + 1; i++) {
    cin >> s[i];
    int v;
    scanf("%d", &v);
    add(v, i);
  }
  int m;
  scanf("%d", &m);
  for (int i = 1; i < m + 1; i++) {
    int v, k;
    scanf("%d%d", &v, &k);
    q[v].push_back(make_pair(k, i));
  }
  deep[0] = 0;
  dfs1(0);
  dfs(0, 1);
  for (int i = 1; i < m + 1; i++) printf("%d\n", ans[i]);
  return 0;
}