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 N = 1e6 + 5;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return f * x;
}
struct node {
  int x;
  int y;
};
string s[N];
vector<node> V[N];
int ver[N], edge[N], head[N], nexts[N], e = 0;
inline void add(int x, int y) {
  ver[++e] = y;
  nexts[e] = head[x];
  head[x] = e;
}
int dep[N];
map<string, int> num[N];
int size[N], son[N], deep[N], fa[N], sum[N];
void dfs(int x) {
  dep[x] = dep[fa[x]] + 1;
  size[x] = 1;
  for (int i = head[x]; i; i = nexts[i]) {
    int v = ver[i];
    dfs(v);
    size[x] += size[v];
    if (size[son[x]] < size[v]) son[x] = v;
  }
}
int skip;
void get(int x, int t) {
  if (num[dep[x]][s[x]] == 0) {
    num[dep[x]][s[x]] += t;
    sum[dep[x]] += t;
  } else {
    num[dep[x]][s[x]] += t;
  }
  if (num[dep[x]][s[x]] == 0) sum[dep[x]] += t;
  for (int i = head[x]; i; i = nexts[i]) {
    int v = ver[i];
    if (skip == v) continue;
    get(v, t);
  }
}
int ans[N];
void make(int x, int op) {
  for (int i = head[x]; i; i = nexts[i]) {
    int v = ver[i];
    if (v == son[x]) continue;
    make(v, 1);
  }
  if (son[x]) make(son[x], 0), skip = son[x];
  get(x, 1);
  skip = 0;
  for (int i = 0; i < V[x].size(); i++)
    ans[V[x][i].x] = sum[dep[x] + V[x][i].y];
  if (op) get(x, -1);
}
int n, m;
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
    fa[i] = read();
    if (fa[i]) add(fa[i], i);
  }
  m = read();
  for (int i = 1; i <= m; i++) {
    int x = read(), y = read();
    V[x].push_back((node){i, y});
  }
  for (int i = 1; i <= n; i++) {
    if (fa[i] == 0) {
      dfs(i);
      make(i, 1);
    }
  }
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}
