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;
int read() {
  char c;
  int w = 1;
  while ((c = getchar()) > '9' || c < '0')
    if (c == '-') w = -1;
  int ans = c - '0';
  while ((c = getchar()) >= '0' && c <= '9')
    ans = (ans << 1) + (ans << 3) + c - '0';
  return ans * w;
}
int n, m;
const int xx = 5e5 + 5;
struct node {
  int next, to;
} e[xx];
int cnt;
int h[xx];
void add(int x, int y) {
  cnt++;
  e[cnt].next = h[x];
  h[x] = cnt;
  e[cnt].to = y;
}
char s[xx];
int dep[xx];
int size[xx];
int son[xx];
void dfs(int x, int y) {
  dep[x] = dep[y] + 1;
  size[x] = 1;
  int maxx = -1;
  for (int i = h[x]; i; i = e[i].next) {
    dfs(e[i].to, x);
    size[x] += size[e[i].to];
    if (size[e[i].to] > maxx) maxx = size[e[i].to], son[x] = e[i].to;
  }
}
string a[xx];
vector<int> id[xx];
vector<int> v[xx];
int ans[xx];
int res[xx];
map<string, int> mp[xx];
void adds(int x, int op) {
  mp[dep[x]][a[x]] += op;
  if (mp[dep[x]][a[x]] == 1 && op == 1) res[dep[x]]++;
  if (mp[dep[x]][a[x]] == 0 && op == -1) res[dep[x]]--;
  for (int i = h[x]; i; i = e[i].next) adds(e[i].to, op);
}
void dfss(int x) {
  for (int i = h[x]; i; i = e[i].next) {
    if (e[i].to == son[x]) continue;
    dfss(e[i].to);
    adds(e[i].to, -1);
  }
  if (son[x]) dfss(son[x]);
  if (x) {
    mp[dep[x]][a[x]] += 1;
    if (mp[dep[x]][a[x]] == 1) res[dep[x]]++;
  }
  for (int i = h[x]; i; i = e[i].next)
    if (e[i].to != son[x]) adds(e[i].to, 1);
  for (int i = 0; i < id[x].size(); i++) ans[id[x][i]] = res[v[x][i]];
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    scanf("%s", s);
    a[i] = s;
    add(read(), i);
  }
  dfs(0, 0);
  m = read();
  for (int i = 1; i <= m; i++) {
    int a, b;
    a = read();
    b = read();
    id[a].push_back(i);
    v[a].push_back(dep[a] + b);
  }
  dfss(0);
  for (int i = 1; i <= m; i++) cout << ans[i] << "\n";
  return 0;
}
