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;
template <typename Tp>
void read(Tp &x) {
  x = 0;
  int fh = 1;
  char ch = 1;
  while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-') fh = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  x *= fh;
}
template <typename Tp>
void biread(Tp &x) {
  x = 0;
  int fh = 1;
  char ch = 1;
  while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-') fh = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 2 + ch - '0', ch = getchar();
  x *= fh;
}
const int maxn = 100000 + 7;
const int maxm = 200000 + 7;
int Head[maxn], Next[maxm], to[maxm], tot, a[maxn], ind, ans[maxn];
void addedge(int x, int y) {
  to[++tot] = y, Next[tot] = Head[x], Head[x] = tot;
}
void add(int x, int y) {
  addedge(x, y);
  addedge(y, x);
}
map<string, int> mp;
int n, T;
struct Question {
  int id, dep;
};
vector<Question> v[maxn];
int size[maxn], dep[maxn], dfn[maxn], son[maxn], cnt, Pre[maxn];
void dfs1(int x, int f, int dp) {
  size[x] = 1, dep[x] = dp, dfn[x] = ++cnt, Pre[cnt] = x;
  int mx = -1;
  for (int i = Head[x]; i; i = Next[i]) {
    int y = to[i];
    if (y == f) continue;
    dfs1(y, x, dp + 1);
    size[x] += size[y];
    if (size[y] > mx) mx = size[y], son[x] = y;
  }
}
int curr[maxn];
map<pair<int, int>, int> has;
void Add(int x, int opt) {
  int L1 = dfn[x], R2 = dfn[x] + size[x] - 1;
  int L2, R1;
  if (!son[x] || opt == -1)
    R1 = R2, R2 = 0, L2 = 1;
  else
    R1 = dfn[son[x]] - 1, L2 = dfn[son[x]] + size[son[x]];
  for (int i = L1; i <= R1; i++) {
    has[make_pair(a[Pre[i]], dep[Pre[i]])] += opt;
    if ((has[make_pair(a[Pre[i]], dep[Pre[i]])] == 1 && opt == 1) ||
        (has[make_pair(a[Pre[i]], dep[Pre[i]])] == 0 && opt == -1)) {
      curr[dep[Pre[i]]] += opt;
    }
  }
  for (int i = L2; i <= R2; i++) {
    has[make_pair(a[Pre[i]], dep[Pre[i]])] += opt;
    if ((has[make_pair(a[Pre[i]], dep[Pre[i]])] == 1 && opt == 1) ||
        (has[make_pair(a[Pre[i]], dep[Pre[i]])] == 0 && opt == -1)) {
      curr[dep[Pre[i]]] += opt;
    }
  }
}
void dfs2(int x, int f, bool op) {
  for (int i = Head[x]; i; i = Next[i]) {
    int y = to[i];
    if (y == f || y == son[x]) continue;
    dfs2(y, x, false);
  }
  if (son[x]) dfs2(son[x], x, true);
  Add(x, 1);
  for (auto p : v[x]) {
    int id = p.id, dp = p.dep + dep[x];
    if (dp > n + 1)
      ans[id] = 0;
    else
      ans[id] = curr[dp];
  }
  if (op == false) Add(x, -1);
}
vector<int> rt;
void Init(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  string s;
  for (int i = 1, p; i <= n; i++) {
    cin >> s >> p;
    if (p)
      add(i, p);
    else
      rt.push_back(i);
    if (mp[s] == 0) mp[s] = ++ind;
    a[i] = mp[s];
  }
  for (auto x : rt) add(x, n + 1);
  cin >> T;
  for (int i = 1, x, y; i <= T; i++) {
    cin >> x >> y;
    v[x].push_back((Question){i, y});
  }
}
void Work(void) {
  dfs1(n + 1, 0, 1);
  dfs2(n + 1, 0, true);
  for (int i = 1; i <= T; i++) cout << ans[i] << endl;
}
signed main(void) {
  Init();
  Work();
  return 0;
}
