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 maxn = 100010;
const int logn = 20;
vector<int> g[maxn];
int parent[maxn], P[logn][maxn], n;
int st[maxn], en[maxn], euler[maxn], depth[maxn], ttt = 1;
void dfs(int u, int par = 0, int dep = 1) {
  parent[u] = par;
  depth[u] = dep;
  st[u] = ttt;
  for (auto first : g[u]) {
    if (first != par) {
      dfs(first, u, dep + 1);
    }
  }
  euler[ttt] = dep;
  en[u] = ttt++;
}
void sparse() {
  for (int i = 1; i <= n; ++i) P[0][i] = parent[i];
  for (int j = 1; j < logn; ++j) {
    for (int i = 1; i <= n; ++i) {
      P[j][i] = P[j - 1][P[j - 1][i]];
    }
  }
}
int kthParent(int n, int k) {
  for (int i = 0; (1 << i) <= k; i++) {
    if (k & (1 << i)) {
      n = P[i][n];
    }
  }
  return n;
}
int s[maxn], tree[logn][maxn], L[logn][maxn];
long long ls[logn][maxn], SL;
void build(int l, int r, int d) {
  if (l == r) return;
  int mid = (l + r) >> 1, cnt = 0, lc = l, rc = mid + 1, ec = 0;
  for (int i = l; i <= r; i++)
    if (tree[d][i] < s[mid]) cnt++;
  for (int i = l; i <= r; i++) {
    if ((tree[d][i] < s[mid]) ||
        (tree[d][i] == s[mid] && ec < (mid - l + 1 - cnt))) {
      tree[d + 1][lc++] = tree[d][i];
      ls[d][i] = ls[d][i - 1] + tree[d][i];
      if (tree[d][i] == s[mid]) ec++;
    } else {
      tree[d + 1][rc++] = tree[d][i];
      ls[d][i] = ls[d][i - 1];
    }
    L[d][i] = L[d][l - 1] + lc - l;
  }
  build(l, mid, d + 1);
  build(mid + 1, r, d + 1);
}
void init(int *arr, int n) {
  for (int i = 1; i <= n; i++) {
    tree[0][i] = arr[i];
    s[i] = tree[0][i];
  }
  sort(s + 1, s + n + 1);
  build(1, n, 0);
}
int LTE(int l, int r, int d, int first, int second, int k) {
  if (l == r) {
    if (l > second || l < first) return 0;
    SL += tree[d][l] * (tree[d][l] <= k);
    return tree[d][l] <= k;
  }
  int cnt = L[d][second] - L[d][first - 1], mid = (l + r) >> 1;
  if (s[mid] <= k) {
    int newx = first - l - (L[d][first - 1] - L[d][l - 1]);
    int newy = second - l + 1 - (L[d][second] - L[d][l - 1]);
    SL += ls[d][second] - ls[d][first - 1];
    return cnt + LTE(mid + 1, r, d + 1, mid + 1 + newx, mid + newy, k);
  } else {
    int newx = L[d][first - 1] - L[d][l - 1];
    int newy = L[d][second] - L[d][l - 1];
    return LTE(l, mid, d + 1, l + newx, l + newy - 1, k);
  }
}
int equal_val(int l, int r, int k) {
  return LTE(1, n, 0, l, r, k) - LTE(1, n, 0, l, r, k - 1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int a;
    cin >> a;
    if (a) {
      g[i].push_back(a);
      g[a].push_back(i);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (!parent[i]) {
      dfs(i);
    }
  }
  sparse();
  init(euler, n);
  int q;
  cin >> q;
  while (q--) {
    int u, k;
    cin >> u >> k;
    u = kthParent(u, k);
    if (u == 0) {
      cout << 0 << ' ';
    } else {
      cout << equal_val(st[u], en[u], k + depth[u]) - 1 << ' ';
    }
  }
  return 0;
}
