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 = 100001;
vector<int> g[N], d[N];
unordered_map<long long, int> f[N];
int i, j, n, m, x, len, y;
int sz[N], h[N], v[N], ans[N], qb[N], cur[N], Fa[N];
char a[N];
long long name[N];
long long cal() {
  long long MOD1 = 100000007, MOD2 = 100000009, p1 = 0, p2 = 0;
  for (int i = 0; i < len; i++)
    p1 = (p1 * 23 + a[i]) % MOD1, p2 = (p2 * 29 + a[i]) % MOD2;
  return p1 * 1000000000 + p2;
}
void init() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%s %d", a, &Fa[i]);
    len = strlen(a);
    x = Fa[i];
    if (x == 0) continue;
    g[x].push_back(i);
    g[i].push_back(x);
    name[i] = cal();
  }
  scanf("%d", &m);
  for (i = 1; i <= m; i++) scanf("%d%d", &x, &qb[i]), d[x].push_back(i);
}
void pre(int x, int fa) {
  sz[x] = 1;
  for (auto y : g[x])
    if (y != fa) {
      h[y] = h[x] + 1;
      pre(y, x);
      sz[x] += sz[y];
    }
}
void add(int x, int fa, int t) {
  if (f[h[x]][name[x]] == 0 && t == 1) cur[h[x]]++;
  if (f[h[x]][name[x]] == 1 && t == -1) cur[h[x]]--;
  f[h[x]][name[x]] += t;
  for (auto y : g[x])
    if (y != fa && !v[y]) add(y, x, t);
}
void dfs(int x, int fa, int keep) {
  int mx = 0, big = -1;
  for (auto y : g[x])
    if (y != fa && sz[y] > mx) mx = sz[y], big = y;
  for (auto y : g[x])
    if (y != fa && y != big) dfs(y, x, 0);
  if (big + 1) dfs(big, x, 1), v[big] = 1;
  add(x, fa, 1);
  for (auto y : d[x])
    if (h[x] + qb[y] < n) ans[y] = cur[h[x] + qb[y]];
  if (big + 1) v[big] = 0;
  if (!keep) add(x, fa, -1);
}
void work() {
  for (i = 1; i <= n; i++)
    if (!Fa[i]) pre(i, 0), dfs(i, 0, 0);
  for (i = 1; i <= m; i++) printf("%d\n", ans[i]);
}
int main() {
  init();
  work();
  return 0;
}
