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 maxn = 1e5 + 5;
int n, m;
vector<int> e[maxn];
vector<string> alls;
string name[maxn];
unordered_map<string, int> vis;
int in[maxn], out[maxn], id[maxn], times;
int sz[maxn], ans[maxn], depth[maxn], son[maxn];
struct Opt {
  int k, id;
};
vector<Opt> opt[maxn];
unordered_map<int, bool> visit[maxn];
int cnt[maxn], col[maxn], mx;
void dfs1(int u, int fa) {
  mx = max(mx, depth[u]);
  in[u] = ++times;
  id[times] = u;
  sz[u] = 1;
  for (auto v : e[u]) {
    depth[v] = depth[u] + 1;
    dfs1(v, u);
    sz[u] += sz[v];
    if (sz[v] > sz[son[u]]) son[u] = v;
  }
  out[u] = times;
}
void dfs2(int u, int fa, int kp) {
  for (auto v : e[u]) {
    if (v == son[u]) continue;
    dfs2(v, u, 0);
  }
  if (son[u] != -1) dfs2(son[u], u, 1);
  if (!visit[depth[u]][col[u]]) visit[depth[u]][col[u]] = 1, cnt[depth[u]]++;
  for (auto v : e[u]) {
    if (v == son[u]) continue;
    for (int i = in[v]; i <= out[v]; ++i) {
      int vv = id[i];
      if (!visit[depth[vv]][col[vv]])
        visit[depth[vv]][col[vv]] = 1, cnt[depth[vv]]++;
    }
  }
  for (auto x : opt[u]) ans[x.id] = cnt[x.k + depth[u]];
  if (!kp) {
    for (int i = in[u]; i <= out[u]; ++i) {
      int v = id[i];
      visit[depth[v]].clear();
      cnt[depth[v]] = 0;
    }
  }
}
int main() {
  scanf("%d", &n);
  int color = 0;
  for (int v = 1; v <= n; ++v) {
    int u;
    cin >> name[v] >> u;
    e[u].push_back(v);
    if (!vis[name[v]])
      col[v] = ++color, vis[name[v]] = color;
    else
      col[v] = vis[name[v]];
  }
  memset(son, -1, sizeof(son));
  dfs1(0, 0);
  scanf("%d", &m);
  for (int i = 1; i <= m; ++i) {
    int u, k;
    scanf("%d%d", &u, &k);
    if (depth[u] + k <= mx) opt[u].push_back({k, i});
  }
  for (auto v : e[0]) dfs2(v, 0, 0);
  for (int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
  return 0;
}
