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 MaxNode = Maxn * 50;
struct Edge {
  int to, nxt;
} edge[Maxn];
struct Treenode {
  int l, r;
  int cnt;
} node[MaxNode];
int Mtail;
int newnode() {
  int x = ++Mtail;
  node[x].l = node[x].r = node[x].cnt = 0;
  return x;
}
int anc[18][Maxn];
int ans[Maxn], dep[Maxn];
int root[Maxn];
int first[Maxn];
int n, Q, te;
vector<pair<int, int> > V[Maxn];
void adde(int x, int y) {
  edge[++te] = Edge{y, first[x]};
  first[x] = te;
}
void update(int x) { node[x].cnt = node[node[x].l].cnt + node[node[x].r].cnt; }
void insert(int &x, int ll, int rr, int p) {
  if (!x) x = newnode();
  if (ll == rr) {
    node[x].cnt++;
    return;
  }
  int mid = ll + rr >> 1;
  if (p <= mid)
    insert(node[x].l, ll, mid, p);
  else
    insert(node[x].r, mid + 1, rr, p);
  update(x);
}
int merge(int x, int y, int ll, int rr) {
  if (!x || !y) return x | y;
  if (ll == rr) {
    node[x].cnt += node[y].cnt;
    return x;
  }
  int mid = ll + rr >> 1;
  node[x].l = merge(node[x].l, node[y].l, ll, mid);
  node[x].r = merge(node[x].r, node[y].r, mid + 1, rr);
  update(x);
  return x;
}
int query(int x, int ll, int rr, int p) {
  if (!x) return 0;
  if (ll == rr) return node[x].cnt;
  int mid = ll + rr >> 1;
  if (p <= mid)
    return query(node[x].l, ll, mid, p);
  else
    return query(node[x].r, mid + 1, rr, p);
}
void dfs(int x) {
  dep[x] = dep[anc[0][x]] + 1;
  insert(root[x], 1, n, dep[x]);
  for (int i = first[x]; i; i = edge[i].nxt) {
    int to = edge[i].to;
    dfs(to);
    root[x] = merge(root[to], root[x], 1, n);
  }
  for (int i = 0; i < V[x].size(); i++)
    ans[V[x][i].second] = query(root[x], 1, n, dep[x] + V[x][i].first) - 1;
}
int getanc(int x, int K) {
  for (int k = 17; k >= 0; k--)
    if (K >= 1 << k) K -= 1 << k, x = anc[k][x];
  return x;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> anc[0][i], adde(anc[0][i], i);
  for (int k = 1; k <= 17; k++)
    for (int i = 1; i <= n; i++) anc[k][i] = anc[k - 1][anc[k - 1][i]];
  cin >> Q;
  for (int x, K, fa, i = 1; i <= Q; i++) {
    cin >> x >> K;
    fa = getanc(x, K);
    V[fa].push_back(make_pair(K, i));
  }
  for (int i = 1; i <= n; i++)
    if (anc[0][i] == 0) dfs(i);
  for (int i = 1; i <= Q; i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}
