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;
template <typename T1, typename T2>
void chkmin(T1 &x, T2 y) {
  if (x > y) x = y;
}
template <typename T1, typename T2>
void chkmax(T1 &x, T2 y) {
  if (x < y) x = y;
}
template <typename T>
void read(T &x) {
  x = 0;
  char c = getchar();
  T neg = 1;
  while (!isdigit(c)) {
    if (c == '-') neg = -1;
    c = getchar();
  }
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  x *= neg;
}
const int MAXN = 2e5;
int n, qu, c[MAXN + 5], rt, idnum = 0;
map<string, int> id;
int hd[MAXN + 5], to[MAXN + 5], nxt[MAXN + 5], ec;
void adde(int u, int v) {
  to[++ec] = v;
  nxt[ec] = hd[u];
  hd[u] = ec;
}
struct query {
  int x, k, id;
} q[MAXN + 5];
vector<int> qv[MAXN + 5];
int siz[MAXN + 5], wson[MAXN + 5], dep[MAXN + 5];
void dfs0(int x, int f) {
  siz[x] = 1;
  for (int e = hd[x]; e; e = nxt[e]) {
    int y = to[e];
    if (y == f) continue;
    dep[y] = dep[x] + 1;
    dfs0(y, x);
    siz[x] += siz[y];
    if (siz[y] > siz[wson[x]]) wson[x] = y;
  }
}
set<int> st[MAXN + 5];
int ans[MAXN + 5];
void del(int x, int f) {
  if (st[dep[x]].find(c[x]) != st[dep[x]].end())
    st[dep[x]].erase(st[dep[x]].find(c[x]));
  for (int e = hd[x]; e; e = nxt[e]) {
    int y = to[e];
    if (y == f) continue;
    del(y, x);
  }
}
void add(int x, int f) {
  st[dep[x]].insert(c[x]);
  for (int e = hd[x]; e; e = nxt[e]) {
    int y = to[e];
    if (y == f) continue;
    add(y, x);
  }
}
void dfs(int x, int f) {
  for (int e = hd[x]; e; e = nxt[e]) {
    int y = to[e];
    if (y == f || y == wson[x]) continue;
    dfs(y, x);
    del(y, x);
  }
  if (wson[x]) dfs(wson[x], x);
  st[dep[x]].insert(c[x]);
  for (int e = hd[x]; e; e = nxt[e]) {
    int y = to[e];
    if (y == f || y == wson[x]) continue;
    add(y, x);
  }
  for (__typeof(qv[x].begin()) it = qv[x].begin(); it != qv[x].end(); it++) {
    int id = *it;
    ans[q[id].id] = st[dep[x] + q[id].k].size();
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    string s;
    int f;
    cin >> s >> f;
    if (!id[s]) id[s] = ++idnum;
    c[i] = id[s];
    adde(f, i);
  }
  scanf("%d", &qu);
  for (int i = 1; i <= qu; i++) {
    scanf("%d%d", &q[i].x, &q[i].k);
    q[i].id = i;
    qv[q[i].x].push_back(i);
  }
  dfs0(0, 0);
  dfs(0, 0);
  for (int i = 1; i <= qu; i++) printf("%d\n", ans[i]);
  return 0;
}