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;
string S[100005];
vector<int> V[100005];
int as[200005], q[200005];
struct tree {
  set<string> T[200005];
  int hd[100005], vr[200005], nt[200005], tt;
  int sz[100005], bg[100005], vs[100005], de[100005], mx[100005];
  void push(int x, int y) {
    vr[++tt] = y;
    nt[tt] = hd[x];
    hd[x] = tt;
  }
  void dfs1(int nw, int en) {
    mx[nw] = de[nw] = de[en] + 1;
    sz[nw] = 1;
    for (int i = hd[nw]; i; i = nt[i]) {
      if (vr[i] == en) continue;
      dfs1(vr[i], nw);
      sz[nw] += sz[vr[i]];
      mx[nw] = max(mx[nw], mx[vr[i]]);
      if (bg[nw] == 0 || sz[bg[nw]] < sz[vr[i]]) bg[nw] = vr[i];
    }
    vs[bg[nw]] = 1;
  }
  void dfs2(int nw, int en, int op) {
    if (op) {
      for (int i = hd[nw]; i; i = nt[i]) {
        if (vr[i] == en || vr[i] == bg[nw]) continue;
        dfs2(vr[i], nw, 1);
      }
    }
    if (bg[nw]) dfs2(bg[nw], nw, op);
    for (int i = hd[nw]; i; i = nt[i]) {
      if (vr[i] == en || vr[i] == bg[nw]) continue;
      dfs2(vr[i], nw, 0);
    }
    T[de[nw]].insert(S[nw]);
    if (op) {
      for (int i = (0); i < (V[nw].size()); ++i) {
        as[V[nw][i]] = T[q[V[nw][i]] + de[nw]].size();
      }
    }
    if (vs[nw] || op == 0) return;
    for (int i = (de[nw]); i <= (mx[nw]); ++i) T[i].clear();
  }
} T;
int n, fa[100005], rt, m, x, y;
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n); ++i) {
    cin >> S[i];
    scanf("%d", &fa[i]);
    if (fa[i]) T.push(fa[i], i);
  }
  scanf("%d", &m);
  for (int i = (1); i <= (m); ++i) {
    scanf("%d%d", &x, &y);
    q[i] = y;
    V[x].push_back(i);
  }
  for (int i = (1); i <= (n); ++i) {
    if (fa[i]) continue;
    T.dfs1(i, 0);
    T.dfs2(i, 0, 1);
  }
  for (int i = (1); i <= (m); ++i) printf("%d\n", as[i]);
  return 0;
}
