Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 7, maxm = 1e5 + 7;
inline long long read() {
  long long s = 0, w = 1;
  char ch = getchar();
  while (ch < 48 || ch > 57) {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= 48 && ch <= 57)
    s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();
  return s * w;
}
int head[maxn], Next[maxm], to[maxm], tot;
void add(int x, int y) {
  to[++tot] = y;
  Next[tot] = head[x];
  head[x] = tot;
}
int size[maxn], son[maxn], dep[maxn];
int cnt[maxn], ans[maxn], dp[maxn][20];
bool vis[maxn];
int que[maxn];
vector<pair<int, int> > query[maxn];
char val[maxn];
void dfs1(int first, int f) {
  size[first] = 1;
  dep[first] = dep[f] + 1;
  dp[first][0] = f;
  for (int i = 1; i < 18; ++i)
    if (dp[first][i - 1] != -1) dp[first][i] = dp[dp[first][i - 1]][i - 1];
  for (int i = head[first], v; i; i = Next[i]) {
    v = to[i];
    dfs1(v, first);
    size[first] += size[v];
    if (size[son[first]] < size[v]) son[first] = v;
  }
}
void calc(int first) {
  cnt[dep[first]] += 1;
  for (int i = head[first], v; i; i = Next[i]) {
    v = to[i];
    if (vis[v]) continue;
    calc(v);
  }
}
bool check(int a[]) {
  int sum = 0;
  for (int i = 0; i < 26; ++i)
    if (a[i] & 1) ++sum;
  return sum <= 1;
}
void delet(int first) {
  cnt[dep[first]] -= 1;
  for (int i = head[first], v; i; i = Next[i]) {
    v = to[i];
    delet(v);
  }
}
void dfs2(int first, bool opt) {
  for (int i = head[first], v; i; i = Next[i]) {
    v = to[i];
    if (son[first] == v) continue;
    dfs2(v, false);
  }
  if (son[first]) {
    dfs2(son[first], true);
    vis[son[first]] = true;
  }
  calc(first);
  int size = query[first].size();
  if (size) {
    for (int i = 0; i < size; ++i)
      ans[query[first][i].first] = cnt[query[first][i].second] - 1;
  }
  if (son[first]) vis[son[first]] = false;
  if (!opt) delet(first);
}
int main() {
  int n = read(), top = 0;
  memset(dp, -1, sizeof dp);
  for (int i = 1, u; i <= n; ++i) {
    u = read();
    if (u)
      add(u, i);
    else
      que[++top] = i;
  }
  int m = read();
  for (int i = 1; i <= top; ++i) dfs1(que[i], 0);
  for (int i = 1, a, k, tmp; i <= m; ++i) {
    a = read(), k = read();
    for (int i = 17; i >= 0; --i)
      if ((1 << i) & k) a = dp[a][i];
    query[a].push_back(make_pair(i, dep[a] + k));
  }
  for (int i = 1; i <= top; ++i) dfs2(que[i], false);
  for (int i = 1; i <= m; ++i) printf("%d ", ans[i]);
  return 0;
}