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 = 100005;
const int maxm = 200005;
const int inf = 1e9 + 7;
int rt[maxn * 40], ls[maxn * 40], rs[maxn * 40], sz[maxn * 40];
int a[maxn], n, q, ct, fa[maxn][25], dep[maxn], L[maxn], R[maxn];
int id[maxn], rid[maxn], ctmp, pos[maxn], d[maxn], val[maxn];
int head[maxn], to[maxm], nex[maxm], cnt;
void add(int u, int v) {
  to[cnt] = v;
  nex[cnt] = head[u];
  head[u] = cnt++;
}
unordered_map<string, int> mp;
void update(int &now, int las, int l, int r, int pos, int v) {
  now = ++ct;
  ls[now] = ls[las];
  rs[now] = rs[las];
  sz[now] = sz[las] + v;
  if (l == r) return;
  int mid = (l + r) / 2;
  if (pos <= mid)
    update(ls[now], ls[las], l, mid, pos, v);
  else
    update(rs[now], rs[las], mid + 1, r, pos, v);
}
int query(int now, int las, int l, int r, int ql, int qr) {
  if (ql <= l && r <= qr) {
    return sz[now] - sz[las];
  }
  int mid = (l + r) / 2, ret = 0;
  if (ql <= mid) ret += query(ls[now], ls[las], l, mid, ql, qr);
  if (qr > mid) ret += query(rs[now], rs[las], mid + 1, r, ql, qr);
  return ret;
}
int gain(string s) {
  if (!mp[s]) mp[s] = ++ctmp;
  return mp[s];
}
void bfs() {
  queue<int> Q;
  Q.push(1);
  int tmp = 0;
  while (!Q.empty()) {
    int v = Q.front();
    Q.pop();
    id[v] = ++tmp;
    a[tmp] = val[v];
    rid[tmp] = v;
    for (int i = head[v], x = to[i]; ~i; i = nex[i], x = to[i]) {
      Q.push(x);
    }
  }
}
void dfs(int u, int dep) {
  d[u] = dep;
  L[dep] = min(L[dep], id[u]);
  R[dep] = max(R[dep], id[u]);
  for (int i = head[u]; ~i; i = nex[i]) {
    fa[to[i]][0] = u;
    dfs(to[i], dep + 1);
  }
}
void build() {
  for (int j = 1; (1 << j) <= n; j++)
    for (int i = 1; i <= n; i++) fa[i][j] = fa[fa[i][j - 1]][j - 1];
  for (int i = 1; i <= n; i++) {
    update(rt[i], rt[i - 1], 1, n, i, 1);
    if (pos[a[i]]) {
      int tmp = rt[i];
      update(rt[i], tmp, 1, n, pos[a[i]], -1);
    }
    pos[a[i]] = i;
  }
}
int getfa(int u, int k) {
  for (int i = 21; i >= 0; i--) {
    if ((k & (1 << i)) == (1 << i)) {
      u = fa[u][i];
    }
  }
  return u;
}
int main() {
  memset(head, -1, sizeof(head));
  scanf("%d", &n);
  for (int i = (int)2; i <= (int)n + 1; i++) {
    string s;
    int ff;
    cin >> s >> ff;
    int ID = gain(s);
    val[i] = ID;
    ff++;
    add(ff, i);
  }
  n++;
  bfs();
  for (int i = (int)1; i <= (int)n; i++) L[i] = inf, R[i] = 0;
  dfs(1, 1);
  build();
  scanf("%d", &q);
  while (q--) {
    int p, k;
    scanf("%d%d", &p, &k);
    p++;
    int aimd = d[p] + k, l, r, aiml = -1, aimr = -1;
    l = L[aimd], r = R[aimd];
    while (l <= r) {
      int mid = (l + r) / 2;
      int faa = getfa(rid[mid], k);
      if (id[faa] >= id[p])
        aiml = mid, r = mid - 1;
      else
        l = mid + 1;
    }
    l = L[aimd], r = R[aimd];
    while (l <= r) {
      int mid = (l + r) / 2;
      int faa = getfa(rid[mid], k);
      if (id[faa] <= id[p])
        aimr = mid, l = mid + 1;
      else
        r = mid - 1;
    }
    if (getfa(rid[aiml], k) != p || aiml < L[aimd] || aiml > R[aimd] ||
        aimr < L[aimd] || aimr > R[aimd]) {
      printf("0\n");
      continue;
    }
    printf("%d\n", query(rt[aimr], 0, 1, n, aiml, aimr));
  }
  return 0;
}
