246E Blood Cousins Return

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 v_i, k_i. Help him to learn for each pair v_i, k_i the number of distinct names among all names of the k_i-sons of the man with numberv_i.

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 s_i and integer r_i (0 ≤ r_i ≤ n), where s_i is the name of the man with a number i, and r_i 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 v_i, k_i (1 ≤ v_i, k_i ≤ 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.

Sample test(s)
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

分析：将节点按深度用vector分别存储，将递归时每个节点的头位置和末位置记录下来，搜索某节点n的k儿子时，用头、末位置对可以取的该深度的节点作出限制，然后用set保存名字字符串，set可去重。

#include<cstdio>
#include<utility>
#include<map>
#include<cstring>
#include<string>
#include<vector>
#include<set>
#include<algorithm>
#include<iterator>
#include<iostream>
using namespace std;

#define maxn 100010
typedef pair<int, int> pii;


vector<int> adj[maxn];
vector< pii > point[maxn];

map<pii, int> vis;
char s[maxn][23];
int pos[maxn][2], depth[maxn], tot, n, v, k;

void dfs(int k, int d) {
    pos[k][0] = ++tot;
    point[d].push_back(make_pair(tot, k));
    depth[k] = d;
    for (vector<int>::iterator p = adj[k].begin(); p != adj[k].end(); ++p) {
        dfs(*p, d + 1);
    }
    pos[k][1] = ++tot;
}

int solve() {
    if (vis[make_pair(v, k)])
        return vis[make_pair(v, k)] - 1;
    vector<pii>::iterator l, r;
    set <string> str;
    l = lower_bound(point[depth[v] + k].begin(), point[depth[v] + k].end(), make_pair(pos[v][0], maxn));
    r = lower_bound(point[depth[v] + k].begin(), point[depth[v] + k].end(), make_pair(pos[v][1], 0));
    for (vector<pii>::iterator p = l; p != r; ++p) {
        str.insert(s[(*p).second]);
    }
    return (vis[make_pair(v, k)] = str.size() + 1) - 1;
}

int main() {
    int m, i, p;
    scanf("%d", &n);
    for (i = 1; i <= n; ++i) {
        scanf("%s%d", &s[i], &p);
        adj[p].push_back(i);
    }
    dfs(0, 0);
    scanf("%d", &m);
    while (m--) {
        scanf("%d%d", &v, &k);
        printf("%d\n", solve());
    }
    return 0;
}