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 INF = 0x3f3f3f3f;
const long long inf = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 5;
int tot, head[maxn];
struct E {
  int to, next;
} edge[maxn << 1];
void add(int u, int v) {
  edge[tot].to = v;
  edge[tot].next = head[u];
  head[u] = tot++;
}
int n;
map<string, int> mp;
int c[maxn], root[maxn];
vector<pair<int, int> > q[maxn];
int siz[maxn], son[maxn], dep[maxn];
void dfs1(int u, int f) {
  dep[u] = dep[f] + 1;
  siz[u] = 1;
  for (int i = head[u]; i != -1; i = edge[i].next) {
    int v = edge[i].to;
    if (v == f) continue;
    dfs1(v, u);
    siz[u] += siz[v];
    if (siz[v] > siz[son[u]]) son[u] = v;
  }
}
int flag, ans[maxn];
int cnt[maxn];
map<pair<int, int>, int> pk;
void count(int u, int f, int val) {
  if (val == 1 && !pk[{dep[u], c[u]}])
    cnt[dep[u]] += val, pk[{dep[u], c[u]}] += val;
  if (val == -1 && pk[{dep[u], c[u]}])
    cnt[dep[u]] += val, pk[{dep[u], c[u]}] += val;
  for (int i = head[u]; i != -1; i = edge[i].next) {
    int v = edge[i].to;
    if (v == f || v == flag) continue;
    count(v, u, val);
  }
}
void dfs(int u, int f, int keep) {
  for (int i = head[u]; i != -1; i = edge[i].next) {
    int v = edge[i].to;
    if (v == f || v == son[u]) continue;
    dfs(v, u, 0);
  }
  if (son[u]) {
    dfs(son[u], u, 1);
    flag = son[u];
  }
  count(u, f, 1);
  for (int i = 0; i < q[u].size(); i++) {
    int id = q[u][i].first;
    int d = q[u][i].second;
    ans[id] = cnt[dep[u] + d];
  }
  flag = 0;
  if (!keep) {
    count(u, f, -1);
  }
}
int main() {
  cin >> n;
  memset(head, -1, sizeof(head));
  int cntcolor = 0;
  int cntroot = 0;
  for (int i = 1; i <= n; i++) {
    string s;
    int x;
    cin >> s >> x;
    if (!mp[s]) mp[s] = ++cntcolor;
    c[i] = mp[s];
    if (!x) {
      ++cntroot;
      root[cntroot] = i;
    } else {
      add(x, i);
      add(i, x);
    }
  }
  int m;
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int u, d;
    cin >> u >> d;
    q[u].push_back({i, d});
  }
  for (int i = 1; i <= cntroot; i++) {
    dfs1(root[i], 0);
  }
  for (int i = 1; i <= cntroot; i++) {
    dfs(root[i], 0, 0);
  }
  for (int i = 1; i <= m; i++) {
    cout << ans[i] << '\n';
  }
}
