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 to[MAXN << 1], nxt[MAXN << 1], head[MAXN];
int cnte;
int v[MAXN], ans[MAXN];
int N, Q, id;
vector<pair<int, int>> Qr[MAXN];
inline void addedge(int x, int y) {
  to[++cnte] = y;
  nxt[cnte] = head[x];
  head[x] = cnte;
}
struct DSU {
  int sz[MAXN], dep[MAXN], son[MAXN], Sz[MAXN];
  int Son;
  map<int, int> cnt[MAXN];
  inline void Init() { memset(son, -1, sizeof(son)); }
  inline void dfs1(int x, int fth) {
    sz[x] = 1;
    dep[x] = dep[fth] + 1;
    for (int i = head[x]; i; i = nxt[i]) {
      int y = to[i];
      if (y != fth) {
        dfs1(y, x);
        sz[x] += sz[y];
        if (son[x] == -1 || sz[y] > sz[son[x]]) son[x] = y;
      }
    }
  }
  inline void add(int x, int fth, int flg) {
    if (flg == 1) {
      cnt[dep[x]][v[x]]++;
      if (cnt[dep[x]][v[x]] == 1) Sz[dep[x]]++;
    } else {
      cnt[dep[x]][v[x]]--;
      if (!cnt[dep[x]][v[x]]) Sz[dep[x]]--;
    }
    for (int i = head[x]; i; i = nxt[i]) {
      int y = to[i];
      if (y != fth && y != Son) add(y, x, flg);
    }
  }
  inline void dfs2(int x, int fth, int opt) {
    for (int i = head[x]; i; i = nxt[i]) {
      int y = to[i];
      if (y != fth && y != son[x]) dfs2(y, x, 0);
    }
    if (son[x] != -1) {
      dfs2(son[x], x, 1);
      Son = son[x];
    } else
      Son = -1;
    add(x, fth, 1);
    Son = -1;
    for (auto now : Qr[x]) {
      int k = now.first, id = now.second;
      if (dep[x] + k >= MAXN)
        ans[id] = 0;
      else
        ans[id] = (int)Sz[dep[x] + k];
    }
    if (!opt) add(x, fth, -1);
  }
} dsu;
map<string, int> dfn;
string S;
int main() {
  ios::sync_with_stdio(false);
  cin >> N;
  for (int i = 1; i <= N; i++) {
    int fa;
    cin >> S;
    cin >> fa;
    if (!dfn[S]) dfn[S] = ++id;
    v[i] = dfn[S];
    addedge(fa, i);
  }
  cin >> Q;
  for (int i = 1; i <= Q; i++) {
    int x, k;
    cin >> x >> k;
    Qr[x].push_back({k, i});
  }
  dsu.Init();
  dsu.dfs1(0, 0);
  dsu.dfs2(0, 0, 0);
  for (int i = 1; i <= Q; i++) cout << ans[i] << '\n';
  return 0;
}
