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>
#pragma GGG optimize(2)
#pragma GGG optimize(3)
#pragma GGG optimize("Ofast")
#pragma GGG optimize("inline")
#pragma GGG optimize("-fgcse")
#pragma GGG optimize("-fgcse-lm")
#pragma GGG optimize("-fipa-sra")
#pragma GGG optimize("-ftree-pre")
#pragma GGG optimize("-ftree-vrp")
#pragma GGG optimize("-fpeephole2")
#pragma GGG optimize("-ffast-math")
#pragma GGG optimize("-fsched-spec")
#pragma GGG optimize("unroll-loops")
#pragma GGG optimize("-falign-jumps")
#pragma GGG optimize("-falign-loops")
#pragma GGG optimize("-falign-labels")
#pragma GGG optimize("-fdevirtualize")
#pragma GGG optimize("-fcaller-saves")
#pragma GGG optimize("-fcrossjumping")
#pragma GGG optimize("-fthread-jumps")
#pragma GGG optimize("-funroll-loops")
#pragma GGG optimize("-fwhole-program")
#pragma GGG optimize("-freorder-blocks")
#pragma GGG optimize("-fschedule-insns")
#pragma GGG optimize("inline-functions")
#pragma GGG optimize("-ftree-tail-merge")
#pragma GGG optimize("-fschedule-insns2")
#pragma GGG optimize("-fstrict-aliasing")
#pragma GGG optimize("-fstrict-overflow")
#pragma GGG optimize("-falign-functions")
#pragma GGG optimize("-fcse-skip-blocks")
#pragma GGG optimize("-fcse-follow-jumps")
#pragma GGG optimize("-fsched-interblock")
#pragma GGG optimize("-fpartial-inlining")
#pragma GGG optimize("no-stack-protector")
#pragma GGG optimize("-freorder-functions")
#pragma GGG optimize("-findirect-inlining")
#pragma GGG optimize("-fhoist-adjacent-loads")
#pragma GGG optimize("-frerun-cse-after-loop")
#pragma GGG optimize("inline-small-functions")
#pragma GGG optimize("-finline-small-functions")
#pragma GGG optimize("-ftree-switch-conversion")
#pragma GGG optimize("-foptimize-sibling-calls")
#pragma GGG optimize("-fexpensive-optimizations")
#pragma GGG optimize("-funsafe-loop-optimizations")
#pragma GGG optimize("inline-functions-called-once")
#pragma GGG optimize("-fdelete-null-pointer-checks")
using namespace std;
const int N = 1000005;
int n, m;
int u;
int tot, head[N], ver[N], nxt[N];
inline void add(int x, int y) {
  ver[++tot] = y;
  nxt[tot] = head[x];
  head[x] = tot;
}
int depth[N], siz[N], son[N];
inline void dfs1(int p, int pa) {
  depth[p] = depth[pa] + 1;
  siz[p] = 1;
  for (int i = head[p]; i; i = nxt[i]) {
    int to = ver[i];
    dfs1(to, p);
    siz[p] += siz[to];
    if (siz[to] > siz[son[p]]) son[p] = to;
  }
}
map<int, int> tt[N];
set<int> cnt[N];
vector<int> Q[N], id[N];
int qx;
int ans[N];
map<string, int> val;
int vv[N];
int idd;
string name;
inline void dfs(int p, int Son, bool del) {
  if (del) {
    --tt[depth[p]][vv[p]];
    if (!tt[depth[p]][vv[p]]) cnt[depth[p]].erase(vv[p]);
  } else {
    if (!tt[depth[p]][vv[p]]) cnt[depth[p]].insert(vv[p]);
    ++tt[depth[p]][vv[p]];
  }
  for (int i = head[p]; i; i = nxt[i])
    if (ver[i] != Son) dfs(ver[i], Son, del);
}
inline void dsu(int p, bool del) {
  for (int i = head[p]; i; i = nxt[i]) {
    int to = ver[i];
    if (to == son[p]) continue;
    dsu(to, 1);
  }
  if (son[p]) dsu(son[p], 0);
  dfs(p, son[p], 0);
  for (int i = 0; i < Q[p].size(); ++i) ans[id[p][i]] = cnt[Q[p][i]].size();
  if (del) dfs(p, 0, 1);
}
bool rt[N];
inline int read() {
  int a = 0;
  int f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) a = a * 10 + (ch ^ 48), ch = getchar();
  return a * f;
}
inline void write(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) write(x / 10);
  putchar(x % 10 + 48);
}
inline void pc() { putchar('\n'); }
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) {
    cin >> name;
    if (!val.count(name)) val[name] = ++idd;
    vv[i] = val[name];
    u = read();
    if (u)
      add(u, i);
    else
      rt[i] = 1;
  }
  for (int i = 1; i <= n; ++i)
    if (rt[i]) dfs1(i, 0);
  m = read();
  for (int i = 1; i <= m; ++i) {
    qx = read();
    Q[qx].push_back(read() + depth[qx]);
    id[qx].push_back(i);
  }
  for (int i = 1; i <= n; ++i)
    if (rt[i]) dsu(i, 1);
  for (int i = 1; i <= m; ++i) write(ans[i]), pc();
  return 0;
}