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() {
  int x = 0, y = 1;
  char ch = ' ';
  for (; (ch != '-' && (ch > '9' || ch < '0')); ch = getchar())
    ;
  if (ch == '-') y = -1, ch = getchar();
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - 48;
  return x * y;
}
const int N = 1e5 + 10;
struct node {
  int to, nex;
} e[N << 1];
int bgn[N], d[N], c[N], sz[N], son[N], ans[N];
string s[N];
map<string, int> mp[N];
vector<pair<int, int> > v[N];
int n, m, cnt, ban;
void add(int x, int y) {
  e[++cnt] = (node){y, bgn[x]};
  bgn[x] = cnt;
}
void dfs(int x, int fa) {
  sz[x] = 1;
  d[x] = d[fa] + 1;
  for (register int i = bgn[x]; i; i = e[i].nex) {
    int y = e[i].to;
    if (y == fa) continue;
    dfs(y, x);
    sz[x] += sz[y];
    if (sz[y] > sz[son[x]]) son[x] = y;
  }
}
void cal(int x, int fa, int op) {
  mp[d[x]][s[x]] += op;
  if (op < 0 && mp[d[x]][s[x]] == 0)
    --c[d[x]];
  else if (op > 0 && mp[d[x]][s[x]] == 1)
    ++c[d[x]];
  for (register int i = bgn[x]; i; i = e[i].nex) {
    int y = e[i].to;
    if (y == fa || y == ban) continue;
    cal(y, x, op);
  }
}
void solve(int x, int fa, int op) {
  for (register int i = bgn[x]; i; i = e[i].nex) {
    int y = e[i].to;
    if (y == fa || y == son[x]) continue;
    solve(y, x, 1);
  }
  if (son[x]) {
    solve(son[x], x, 0);
    ban = son[x];
  }
  cal(x, fa, 1);
  for (register int i = 0; i < v[x].size(); ++i) {
    int u = v[x][i].first, p = v[x][i].second;
    if (d[x] + u <= n + 1) ans[p] = c[d[x] + u];
  }
  ban = 0;
  if (op) cal(x, fa, -1);
}
int main() {
  int x, y, z, k;
  n = read();
  for (register int i = 1; i <= n; ++i) {
    cin >> s[i];
    x = read();
    if (x == 0) x = n + 1;
    add(x, i);
    add(i, x);
  }
  dfs(n + 1, 0);
  m = read();
  for (register int i = 1; i <= m; ++i) {
    x = read();
    y = read();
    v[x].push_back(make_pair(y, i));
  }
  solve(n + 1, 0, 0);
  for (register int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
  return 0;
}
