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;
inline int setBit(int N, int pos) { return N = N | (1 << pos); }
inline int resetBit(int N, int pos) { return N = N & ~(1 << pos); }
inline bool checkBit(int N, int pos) { return (bool)(N & (1 << pos)); }
const int MAX = 100005;
int n, level[MAX], par[MAX][20], freq[MAX], ans[MAX], subtree[MAX];
vector<int> graph[MAX], root;
bool heavy[MAX];
struct query {
  int h, idx;
};
vector<query> qList[MAX];
int dfs1(int u, int l = 0) {
  level[u] = l;
  subtree[u] = 1;
  for (auto v : graph[u]) {
    par[v][0] = u;
    subtree[u] += dfs1(v, l + 1);
  }
  return subtree[u];
}
void init() {
  int i, j;
  memset(par, -1, sizeof(par));
  for (auto x : root) dfs1(x);
  for (j = 1; (1 << j) < n; j++) {
    for (i = 0; i < n; i++) {
      if (par[i][j - 1] != -1) par[i][j] = par[par[i][j - 1]][j - 1];
    }
  }
}
inline int p(int u, int k) {
  if (k > level[u]) return -1;
  int i, lg = log2(level[u]);
  for (i = lg; i >= 0; i--) {
    if ((1 << i) <= k) {
      u = par[u][i];
      k -= (1 << i);
    }
  }
  return u;
}
void add(int u, int par = -1, int x = 0) {
  freq[level[u]] += x;
  for (auto v : graph[u]) {
    if (v != par && !heavy[v]) add(v, u, x);
  }
}
void dfs(int u, int par = -1, bool keepFlag = false) {
  int mx = -1, heavyChild = -1;
  for (auto v : graph[u]) {
    if (v != par && subtree[v] > mx) {
      mx = subtree[v];
      heavyChild = v;
    }
  }
  for (auto v : graph[u]) {
    if (v != par && v != heavyChild) {
      dfs(v, u, false);
    }
  }
  if (heavyChild != -1) {
    dfs(heavyChild, u, true);
    heavy[heavyChild] = true;
  }
  add(u, par, 1);
  for (auto x : qList[u]) {
    ans[x.idx] = freq[x.h + level[u]] - 1;
  }
  if (heavyChild != -1) heavy[heavyChild] = false;
  if (!keepFlag) add(u, par, -1);
}
int main() {
  int i, x, m, cousin;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &x);
    if (x == 0)
      root.push_back(i);
    else
      graph[x - 1].push_back(i);
  }
  init();
  scanf("%d", &m);
  for (i = 0; i < m; i++) {
    scanf("%d %d", &x, &cousin);
    int here = p(x - 1, cousin);
    if (here != -1) qList[here].push_back({cousin, i});
  }
  for (auto x : root) dfs(x);
  for (i = 0; i < m; i++) printf("%d ", ans[i]);
  puts("");
  return 0;
}
