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;
struct node {
  int dep, id, nxt;
} q[N];
int n, cnt, col[N], dep[N], m, ans[N], head[N], tot, sz[N], son[N], mx;
bool vis[N];
string tmp;
map<string, int> mp;
map<int, int> st[N];
map<int, int>::iterator it;
vector<int> e[N];
inline void add(int x, int k, int i) {
  q[++tot] = (node){k, i, head[x]};
  head[x] = tot;
}
inline void dfs(int x) {
  sz[x] = 1;
  mx = max(mx, dep[x]);
  for (int i = 0; i < (int)e[x].size(); i++) {
    int v = e[x][i];
    dep[v] = dep[x] + 1;
    dfs(v);
    sz[x] += sz[v];
    son[x] = sz[son[x]] > sz[v] ? son[x] : v;
  }
}
inline void update(int x, int v) {
  st[dep[x]][col[x]] += v;
  if (st[dep[x]][col[x]] == 0) {
    it = st[dep[x]].find(col[x]);
    st[dep[x]].erase(it);
  }
  for (int i = 0; i < (int)e[x].size(); i++)
    if (!vis[e[x][i]]) update(e[x][i], v);
}
inline void Dfs(int x, int keep) {
  if (x == 0) return;
  for (int i = 0; i < (int)e[x].size(); i++) {
    int v = e[x][i];
    if (v == son[x]) continue;
    Dfs(v, 0);
  }
  Dfs(son[x], 1);
  vis[son[x]] = 1;
  update(x, 1);
  vis[son[x]] = 0;
  for (int i = head[x]; i; i = q[i].nxt) {
    int now = dep[x] + q[i].dep;
    if (now <= mx) ans[q[i].id] += st[now].size();
  }
  if (!keep) update(x, -1);
}
int main() {
  scanf("%d", &n);
  for (int i = 1, x; i <= n; i++) {
    cin >> tmp;
    scanf("%d\n", &x);
    e[x].push_back(i);
    if (!mp.count(tmp)) mp[tmp] = ++cnt;
    col[i] = mp[tmp];
  }
  scanf("%d", &m);
  for (int i = 1, x, k; i <= m; i++) {
    scanf("%d%d\n", &x, &k);
    add(x, k, i);
  }
  dfs(0);
  for (int i = 0; i < (int)e[0].size(); i++) {
    int v = e[0][i];
    Dfs(v, 0);
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}
