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;
inline void read(int &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9')
    x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  x *= f;
}
const int maxn = 100100;
struct node {
  int v, nxt;
} e[maxn];
int head[maxn], tot;
inline void add(int x, int y) {
  tot++;
  e[tot].v = y;
  e[tot].nxt = head[x];
  head[x] = tot;
}
struct AA {
  int s, num;
};
map<string, int> h;
vector<AA> ask[maxn];
set<int> hip[maxn];
int n, q, s, t, cnt, name[maxn], dep[maxn], son[maxn], size[maxn], ans[maxn];
bool vis[maxn];
char ch[30];
inline void dfs1(int x, int fa) {
  size[x] = 1;
  dep[x] = dep[fa] + 1;
  for (int i = head[x]; i; i = e[i].nxt) {
    int v = e[i].v;
    if (v == fa) continue;
    dfs1(v, x);
    size[x] += size[v];
    if (son[x] == 0 || size[v] > size[son[x]]) son[x] = v;
  }
}
inline void calc(int x, int fa) {
  hip[dep[x]].insert(name[x]);
  for (int i = head[x]; i; i = e[i].nxt) {
    int v = e[i].v;
    if (v == fa || vis[v]) continue;
    calc(v, x);
  }
}
inline void dfs2(int x, int fa, int keep) {
  for (int i = head[x]; i; i = e[i].nxt) {
    int v = e[i].v;
    if (v == fa || v == son[x]) continue;
    dfs2(v, x, 0);
  }
  if (son[x]) dfs2(son[x], x, 1), vis[son[x]] = true;
  calc(x, fa);
  vis[son[x]] = false;
  for (int i = 0; i < ask[x].size(); i++)
    if (dep[x] + ask[x][i].s <= n)
      ans[ask[x][i].num] = hip[dep[x] + ask[x][i].s].size();
  if (!keep)
    for (int i = dep[x]; !hip[i].empty(); i++) hip[i].clear();
}
int main() {
  read(n);
  for (int i = 1; i <= n; i++) {
    scanf("%s", ch);
    if (h.find(ch) == h.end()) h[ch] = ++cnt;
    name[i] = h[ch];
    read(s), add(s, i);
  }
  read(q);
  AA tmp;
  for (int i = 1; i <= q; i++) {
    read(s), read(t);
    tmp.s = t;
    tmp.num = i;
    ask[s].push_back(tmp);
  }
  dep[0] = -1;
  dfs1(0, 0);
  dfs2(0, 0, 1);
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
  return 0;
}
