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 = 100100;
const long long MOD = 1000000007;
const double EBS = 1e-7;
int dX[] = {0, 0, 1, -1};
int dY[] = {1, -1, 0, 0};
long long GCD(long long num, long long ones) {
  return !ones ? num : GCD(ones, num % ones);
}
const int L = 17;
vector<int> adj[N];
vector<int> Root;
vector<int> level[N];
int depth[N], ID[N], rID[N], dp[L + 1][N], currID = 1, n, m, k, WantedNode;
void dfs(int cur, int par) {
  dp[0][cur] = par;
  ID[cur] = currID++;
  rID[ID[cur]] = cur;
  level[depth[cur]].push_back(ID[cur]);
  for (auto x : adj[cur])
    if (x != par) depth[x] = depth[cur] + 1, dfs(x, cur);
}
inline int getKthAncestor(int u, int k) {
  int d = depth[u] - k;
  for (int j = L; j >= 0; --j) {
    int nu = dp[j][u];
    if (!nu) continue;
    if (depth[nu] == d)
      return nu;
    else if (depth[nu] > d)
      u = nu;
  }
  return 0;
}
inline int BSL(int d) {
  int l = 0, r = (level[d].size()) - 1, mid;
  while (l < r) {
    mid = l + (r - l) / 2;
    int node = getKthAncestor(rID[level[d][mid]], k);
    node = ID[node];
    if (node >= WantedNode)
      r = mid;
    else
      l = mid + 1;
  }
  return r;
}
inline int BSR(int d) {
  int l = 0, r = (level[d].size()) - 1, mid;
  while (l < r) {
    mid = l + (r - l + 1) / 2;
    int node = getKthAncestor(rID[level[d][mid]], k);
    node = ID[node];
    if (node <= WantedNode)
      l = mid;
    else
      r = mid - 1;
  }
  return l;
}
int main() {
  scanf("%d", &n);
  int p, v;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &p);
    if (p)
      adj[p].push_back(i + 1), adj[i + 1].push_back(p);
    else
      Root.push_back(i + 1);
  }
  for (auto x : Root) dfs(x, 0);
  for (int j = 1; j <= L; ++j)
    for (int i = 0; i < N; ++i)
      dp[j][i] = (dp[j - 1][i] == -1 ? -1 : dp[j - 1][dp[j - 1][i]]);
  scanf("%d", &m);
  while (m--) {
    scanf("%d", &v);
    scanf("%d", &k);
    WantedNode = getKthAncestor(v, k);
    if (WantedNode) {
      WantedNode = ID[WantedNode];
      printf("%d ", BSR(depth[v]) - BSL(depth[v]));
    } else
      printf("0 ");
  }
  return 0;
}
