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 long long maxn = 2e5 + 10;
int head[maxn << 1], to[maxn << 1], nex[maxn << 1], e;
int fa[maxn], dep[maxn], son[maxn], siz[maxn], vis[maxn];
string a[maxn];
int sum[maxn];
vector<pair<int, int> > q[maxn];
int ans[maxn];
map<string, int> mp[maxn];
void add(int x, int y) {
  to[++e] = y;
  nex[e] = head[x];
  head[x] = e;
}
void dfs1(int x, int f, int deep) {
  siz[x] = 1;
  dep[x] = deep;
  int maxson = -1;
  for (int i = head[x]; i; i = nex[i]) {
    int y = to[i];
    if (y == f) continue;
    dfs1(y, x, deep + 1);
    siz[x] += siz[y];
    if (maxson < siz[y]) {
      maxson = siz[y];
      son[x] = y;
    }
  }
}
void upd(string a, int dep, int val) {
  if (val == 1) {
    ++mp[dep][a];
    if (mp[dep][a] == 1) ++sum[dep];
  } else {
    --mp[dep][a];
    if (mp[dep][a] == 0) --sum[dep];
  }
}
void counts(int x, int val) {
  upd(a[x], dep[x], val);
  for (int i = head[x]; i; i = nex[i]) {
    int y = to[i];
    if (y == fa[x] || vis[y]) continue;
    counts(y, val);
  }
}
void dsu(int x, int f, int hson) {
  for (int i = head[x]; i; i = nex[i]) {
    int y = to[i];
    if (y == f || y == son[x]) continue;
    dsu(y, x, 0);
  }
  if (son[x]) {
    dsu(son[x], x, 1);
    vis[son[x]] = 1;
  }
  counts(x, 1);
  for (int i = 0; i < q[x].size(); i++) {
    int y = q[x][i].second;
    ans[q[x][i].first] = sum[y];
  }
  if (son[x]) vis[son[x]] = 0;
  if (!hson) counts(x, -1);
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i] >> fa[i];
    add(i, fa[i]);
    add(fa[i], i);
  }
  for (int i = 1; i <= n; i++) {
    if (!fa[i]) dfs1(i, 0, 1);
  }
  int m, x, y;
  cin >> m;
  for (int i = 1; i <= m; i++) {
    cin >> x >> y;
    q[x].push_back(make_pair(i, dep[x] + y));
  }
  for (int i = 1; i <= n; i++) {
    if (!fa[i]) dsu(i, 0, 0);
  }
  for (int i = 1; i <= m; i++) {
    printf("%d\n", ans[i]);
  }
}
