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;
vector<int> by[100005];
int n, m;
unsigned int st[100005];
char ss[25];
map<pair<int, int>, int> Map;
int F[100005][20], root = 0;
int Ls[100005], Rs[100005];
vector<int> LevL[100005];
vector<int> LevR[100005];
int ans[100005];
vector<int> lev[100005];
int id[100005];
int dp[100005];
const int Maxint = 0x7fffffff;
unsigned int BKDRHash(char str[]) {
  unsigned int seed = 131;
  unsigned int has = 0;
  for (int i = 0; str[i] != '\0'; i++) has = has * seed + (str[i]);
  return (has & Maxint);
}
void input() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%s%d", ss, &F[i][0]);
    st[i] = BKDRHash(ss);
    by[F[i][0]].push_back(i);
  }
}
void dfs(int rt, int dep, int &step) {
  lev[dep].push_back(rt);
  dp[rt] = dep;
  Ls[rt] = ++step;
  for (int i = 0; i < (int)by[rt].size(); i++) dfs(by[rt][i], dep + 1, step);
  Rs[rt] = ++step;
}
int cmp(int x, int y) { return Ls[x] < y; }
int lowbond(int level, int y) {
  return lower_bound(lev[level].begin(), lev[level].end(), y, cmp) -
         lev[level].begin();
}
void solve() {
  int v, k, l, r;
  scanf("%d", &m);
  memset(ans, 0, sizeof(ans));
  Map.clear();
  int de;
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &v, &k);
    de = dp[v] + k;
    l = lowbond(de, Ls[v]);
    r = lowbond(de, Rs[v]) - 1;
    if (r < l) continue;
    int &t = Map[pair<int, int>(v, k)];
    if (ans[i] = t) continue;
    set<unsigned int> mp;
    for (int j = r; j >= l; j--) {
      int di = lev[de][j];
      mp.insert(st[di]);
    }
    ans[i] = t = mp.size();
  }
  for (int i = 0; i < m; i++) printf("%d\n", ans[i]);
}
int main() {
  input();
  root = 0;
  int step = 0;
  dfs(root, 0, step);
  solve();
  return 0;
}
