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 N = 2e5 + 5;
const int mod = 1e9 + 7;
int n, m;
char str[N];
vector<int> e[N];
int val[N], ans[N], deg[N];
vector<pair<int, int> > que[N];
struct hsh {
  static const int mod1 = 1e9 + 7, mod2 = 1e9 + 9;
  long long a, b;
  hsh(long long a = 0, long long b = 0) : a(a), b(b) {}
  hsh operator+(const hsh& _) const {
    return hsh((a + _.a) % mod1, (b + _.b) % mod2);
  }
  hsh operator-(const hsh& _) const {
    return hsh((a - _.a + mod1) % mod1, (b - _.b + mod2) % mod2);
  }
  hsh operator*(const hsh& _) const {
    return hsh(a * _.a % mod1, b * _.b % mod2);
  }
  bool operator<(const hsh& _) const {
    return a < _.a || (a == _.a && b < _.b);
  }
  bool operator==(const hsh& _) const { return a == _.a && b == _.b; }
  void print() {
    cerr << "a" << '=' << a << ' ', cerr << "b" << '=' << b << endl;
  }
} b[N], a[N];
int cnt;
struct node {
  int dep, col;
  bool operator<(const node& _) const {
    return dep < _.dep || (dep == _.dep && col < _.col);
  }
};
map<node, int> mp[N];
map<int, int> ct[N];
int siz[N], fa[N], dep[N];
int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
void init() {
  for (int i = (0); i < (N); i++) fa[i] = i, siz[i] = 1;
}
void uni(int x, int y) {
  x = find(x), y = find(y);
  if (siz[x] > siz[y]) swap(x, y);
  fa[x] = y;
  siz[y] += siz[x];
  for (map<node, int>::iterator it = mp[x].begin(); it != mp[x].end(); it++) {
    node e1 = (*it).first;
    int num = (*it).second;
    if (mp[y].find(e1) == mp[y].end()) ct[y][e1.dep]++;
    mp[y][e1] += num;
  }
}
void insert(int x, int y) {
  y = find(y);
  siz[y]++;
  fa[x] = y;
  node e1 = (node){dep[x], val[x]};
  ct[y][e1.dep]++;
  mp[y][e1]++;
}
void dfs(int u) {
  int las = 0;
  for (int i = (0); i < ((int)e[u].size()); i++) {
    int v = e[u][i];
    dep[v] = dep[u] + 1;
    dfs(v);
    if (las) uni(las, v);
    las = v;
  }
  if (las)
    insert(u, las);
  else
    mp[u][(node){dep[u], val[u]}] = 1, ct[u][dep[u]] = 1;
  int root = find(u);
  for (int i = (0); i < ((int)que[u].size()); i++) {
    pair<int, int> e1 = que[u][i];
    ans[e1.second] = ct[root][e1.first + dep[u]];
  }
}
int main() {
  init();
  hsh P = hsh(13, 17);
  scanf("%d", &n);
  for (int i = (1); i < (n + 1); i++) {
    int fa;
    scanf("%s%d", str, &fa);
    if (fa) {
      e[fa].push_back(i);
      deg[i]++;
    }
    int len = strlen(str);
    hsh res = hsh(1, 1);
    for (int i = (0); i < (len); i++)
      res = res * P + hsh(str[i] - 'a', str[i] - 'a');
    a[i] = b[++cnt] = res;
  }
  sort(b + 1, b + 1 + cnt);
  cnt = unique(b + 1, b + 1 + cnt) - b - 1;
  for (int i = (1); i < (n + 1); i++)
    val[i] = lower_bound(b + 1, b + 1 + cnt, a[i]) - b;
  scanf("%d", &m);
  for (int i = (1); i < (m + 1); i++) {
    int v, w;
    scanf("%d%d", &v, &w);
    que[v].push_back(make_pair(w, i));
  }
  for (int i = (1); i < (n + 1); i++)
    if (!deg[i]) dfs(i);
  for (int i = (1); i < (m + 1); i++) printf("%d\n", ans[i]);
  return 0;
}
