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;
void rd(int &x) {
  x = 0;
  static char c;
  while (c = getchar(), c < 48)
    ;
  do x = x * 10 + (c ^ 48);
  while (c = getchar(), c > 47);
}
void ptn(int x) {
  if (!x) {
    puts("0");
    return;
  }
  static char stk[35];
  int tp = 0;
  for (; x; x /= 10) stk[tp++] = x % 10;
  for (int i = (tp)-1, _s = (0); i >= _s; --i) putchar(stk[i] ^ 48);
  putchar('\n');
}
const int N = (int)1e5 + 5;
int lst[N], par[N], val[N], dep[N], sz[N], hev[N], dfn[N], post[N], refer[N],
    ans[N], cnt[N], amt[N];
pair<int, int> edge[N];
vector<pair<int, int> > query[N];
void add_edge(int u, int v) {
  static int tot_edge;
  edge[tot_edge] = pair<int, int>(v, lst[u]);
  lst[u] = tot_edge++;
}
void dfs(int u) {
  static int dfs_clock;
  refer[dfn[u] = ++dfs_clock] = u;
  sz[u] = 1;
  for (int i = lst[u]; ~i; i = edge[i].second) {
    int v = edge[i].first;
    dep[v] = dep[u] + 1;
    dfs(v);
    if (!hev[u] || sz[v] > sz[hev[u]]) hev[u] = v;
    sz[u] += sz[v];
  }
  post[u] = dfs_clock;
}
struct Unit {
  int u, Hash;
  Unit() {}
  Unit(int u, int Hash) : u(u), Hash(Hash) {}
  bool operator<(const Unit &rhs) const {
    if (dep[u] != dep[rhs.u]) return dep[u] < dep[rhs.u];
    return Hash < rhs.Hash;
  }
} dat[N];
inline void obtain(int u) {
  for (int i = (dfn[u]), _t = (post[u] + 1); i < _t; ++i)
    if (!cnt[val[refer[i]]]++) ++amt[dep[refer[i]]];
}
inline void remove(int u) {
  for (int i = (dfn[u]), _t = (post[u] + 1); i < _t; ++i)
    if (!--cnt[val[refer[i]]]) --amt[dep[refer[i]]];
}
void redfs(int u) {
  for (int i = lst[u]; ~i; i = edge[i].second) {
    int v = edge[i].first;
    if (v != hev[u]) redfs(v);
  }
  if (hev[u]) redfs(hev[u]);
  for (int i = lst[u]; ~i; i = edge[i].second) {
    int v = edge[i].first;
    if (v != hev[u]) obtain(v);
  }
  if (!cnt[val[u]]++) ++amt[dep[u]];
  for (pair<int, int> elm : query[u]) {
    int tar = dep[u] + elm.second;
    if (tar < N) ans[elm.first] = amt[tar];
  }
  if (u != hev[par[u]]) remove(u);
}
int main() {
  int n;
  rd(n);
  memset(lst, -1, n + 1 << 2);
  for (int i = (1), _t = (n + 1); i < _t; ++i) {
    static char str[25];
    scanf("%s", str), rd(par[i]);
    add_edge(par[i], i);
    int Hash = 0;
    for (int i = (0), _t = (strlen(str)); i < _t; ++i)
      Hash = Hash * 10007 + str[i];
    dat[i] = Unit(i, Hash);
  }
  dfs(0);
  sort(dat + 1, dat + n + 1);
  int allc = 0;
  for (int i = (1), _t = (n + 1); i < _t; ++i) {
    if (dat[i].Hash != dat[i - 1].Hash || dep[dat[i].u] != dep[dat[i - 1].u])
      val[dat[i].u] = ++allc;
    else
      val[dat[i].u] = allc;
  }
  int m;
  rd(m);
  for (int i = (0), _t = (m); i < _t; ++i) {
    static int u, K;
    rd(u), rd(K);
    query[u].push_back(pair<int, int>(i, K));
  }
  redfs(0);
  for (int i = (0), _t = (m); i < _t; ++i) ptn(ans[i]);
  return 0;
}
