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 + 5;
int n, m, tot, Son, col, lnk[maxn], son[maxn], nxt[maxn], maxson[maxn], c[maxn],
    siz[maxn], dep[maxn];
int tot1, lnkq[maxn], sonq[maxn], idq[maxn], nxtq[maxn], ans[maxn];
string s;
map<string, int> hsh1;
map<int, bool> hsh[maxn];
void add(int x, int y) { son[++tot] = y, nxt[tot] = lnk[x], lnk[x] = tot; }
void addq(int x, int y, int z) {
  sonq[++tot1] = y, idq[tot1] = z, nxtq[tot1] = lnkq[x], lnkq[x] = tot1;
}
void dfs1(int x) {
  siz[x] = 1;
  for (int j = lnk[x]; j; j = nxt[j]) {
    dep[son[j]] = dep[x] + 1;
    dfs1(son[j]);
    siz[x] += siz[son[j]];
    if (siz[son[j]] > siz[maxson[x]]) maxson[x] = son[j];
  }
}
void calc(int x, int op) {
  if (op == 1)
    hsh[dep[x]][c[x]] = 1;
  else
    hsh[dep[x]].clear();
  for (int j = lnk[x]; j; j = nxt[j])
    if (son[j] != Son) calc(son[j], op);
}
void dfs(int x, bool pd) {
  for (int j = lnk[x]; j; j = nxt[j])
    if (son[j] != maxson[x]) dfs(son[j], 0);
  if (maxson[x]) dfs(maxson[x], 1), Son = maxson[x];
  calc(x, 1);
  Son = 0;
  for (int j = lnkq[x]; j; j = nxtq[j]) {
    int h = dep[x] + sonq[j];
    if (h > n) continue;
    ans[idq[j]] = hsh[h].size();
  }
  if (!pd) calc(x, -1);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    cin >> s;
    int x;
    scanf("%d", &x);
    add(x, i);
    if (hsh1[s] == 0) hsh1[s] = ++col;
    c[i] = hsh1[s];
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    addq(x, y, i);
  }
  for (int j = lnk[0]; j; j = nxt[j]) dfs1(son[j]);
  for (int j = lnk[0]; j; j = nxt[j]) dfs(son[j], 0);
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}