Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int a[120000];
vector<int> adj[120000];
int vis[120000];
int in_deg[120000];
int level[120000];
int n;
int in_cnt[120000];
int out_cnt[120000];
int lefty[120000];
int rite[120000];
int counter;
void dfs(int u) {
  int i, v;
  vis[u] = 1;
  in_cnt[u] = counter++;
  for (i = 0; i < adj[u].size(); i++) {
    v = adj[u][i];
    if (vis[v]) continue;
    dfs(v);
  }
  out_cnt[u] = counter++;
}
queue<int> q;
void bfs(int u) {
  while (!q.empty()) q.pop();
  int i, v, f;
  memset(level, -1, sizeof(level));
  level[u] = 0;
  q.push(u);
  n = 1;
  while (!q.empty()) {
    f = q.front();
    q.pop();
    for (i = 0; i < adj[f].size(); i++) {
      v = adj[f][i];
      if (level[v] != -1) continue;
      level[v] = level[f] + 1;
      q.push(v);
      a[n++] = v;
    }
  }
  n--;
  for (i = 1; i <= n; i++) {
    if (lefty[level[a[i]]] == 0) {
      lefty[level[a[i]]] = i;
      rite[level[a[i]]] = i;
    } else
      rite[level[a[i]]] = i;
  }
}
char name[200];
int num[100010];
struct Query {
  int l, r;
  int id;
} Q[100010];
struct cmp {
  bool operator()(const Query& q1, const Query& q2) { return q1.l < q2.l; }
};
map<string, int> mp;
vector<int> adj2[110000];
int adj2_id[110000];
int ans[110000];
int tree[110000];
int get(int pos) {
  int ret = 0, i;
  for (i = pos; i; i -= (i & -i)) ret += tree[i];
  return ret;
}
void update(int pos, int val) {
  int i;
  for (i = pos; i < 100010; i += (i & -i)) tree[i] += val;
}
int main() {
  int i, j, k, I = 1, val, l, v, m;
  int lo, hi, lo1, hi1, mid, t_in, t_out;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%s %d", name, &val);
    if (mp[(string)name] == 0) {
      mp[(string)name] = I++;
    }
    num[i] = mp[(string)name];
    if (val == 0) {
    } else {
      in_deg[i] = 1;
      adj[val].push_back(i);
    }
  }
  mp.clear();
  for (i = 1; i <= n; i++) {
    if (in_deg[i] == 0) {
      adj[0].push_back(i);
    }
  }
  dfs(0);
  bfs(0);
  scanf("%d", &m);
  for (i = 1; i <= m; i++) {
    scanf("%d %d", &v, &k);
    l = level[v] + k;
    lo1 = lefty[l];
    hi1 = rite[l];
    lo = lo1;
    hi = hi1;
    t_in = in_cnt[v];
    while (lo <= hi) {
      mid = (lo + hi) / 2;
      if (in_cnt[a[mid]] < t_in)
        lo = mid + 1;
      else
        hi = mid - 1;
    }
    Q[i].l = lo;
    lo = lo1;
    hi = hi1;
    t_out = out_cnt[v];
    while (lo <= hi) {
      mid = (lo + hi) / 2;
      if (out_cnt[a[mid]] > t_out)
        hi = mid - 1;
      else
        lo = mid + 1;
    }
    Q[i].r = hi;
    Q[i].id = i;
  }
  sort(Q + 1, Q + m + 1, cmp());
  for (i = 1; i <= n; i++) {
    adj2[num[a[i]]].push_back(i);
    if (adj2[num[a[i]]].size() == 1) {
      update(i, 1);
      adj2_id[num[a[i]]]++;
    }
  }
  int last = 1;
  for (i = 1; i <= m; i++) {
    if (Q[i].l > Q[i].r) {
      ans[Q[i].id] = 0;
      continue;
    }
    while (last < Q[i].l) {
      if ((get(last) - get(last - 1))) {
        update(last, -1);
        v = num[a[last]];
        if (adj2_id[v] < adj2[v].size()) {
          update(adj2[v][adj2_id[v]], 1);
          adj2_id[v]++;
        }
      }
      last++;
    }
    ans[Q[i].id] = get(Q[i].r) - get(Q[i].l - 1);
  }
  for (i = 1; i <= m; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}