[cf contest246] E - Blood Cousins Return

[cf contest246] E - Blood Cousins Return

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

6pasha 0gerald 1gerald 1valera 2igor 3olesya 151 11 21 33 16 1

output

22010

input

6valera 0valera 1valera 1gerald 0valera 4kolya 471 11 22 12 24 15 16 1

output

1000200

题目大意：给定一棵树，给一些询问询问一个节点x子树里，与x距离为k（k不同）节点的值有多少种。

这里介绍一下dsu on tree这种东西。

这个方法是一种启发式的合并，一般与树剖合用。

思想是暴力思想的优化，与莫队思想类似。而时间复杂度的证明与轻重链（树链剖分）有关。

我先留坑吧。先放一个piano写的贴子。piano

声明一下dsu是一个离线算法，总复杂度一般是O(nlogn)的。要么就乘上一个统计的复杂度。

对于这题，我们主要思考如何统计?

用n个set维护一下就好了。复杂度是O(nlog²n)的。

code：

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

 ;
 int n,q,co[N];
 int tot,lnk[N],nxt[N],son[N];
 int fa[N],dep[N],siz[N],got[N];
 int skp,cnt[N],ans[N];
 map <string,int> rel;
 set <int> ase[N];
 vector <pair <int,int> > a[N];
 void add (int x,int y) {
     nxt[++tot]=lnk[x],lnk[x]=tot;
     son[tot]=y;
 }
 namespace tcd {
     void dfs_ppw (int x,int p) {
         fa[x]=p,dep[x]=dep[p]+;
         siz[x]=,got[x]=;
         for (int j=lnk[x]; j; j=nxt[j]) {
             if (son[j]==p) continue;
             dfs_ppw(son[j],x);
             siz[x]+=siz[son[j]];
             if (siz[son[j]]>siz[got[x]]) got[x]=son[j];
         }
     }
     void calc (int x,int v) {
         &&x>) ase[dep[x]].insert(co[x]);
         else ase[dep[x]].erase(co[x]);
         for (int j=lnk[x]; j; j=nxt[j]) {
             if (son[j]==fa[x]||son[j]==skp) continue;
             calc(son[j],v);
         }
     }
     void main (int x,bool f) {
         for (int j=lnk[x]; j; j=nxt[j]) {
             if (son[j]==fa[x]||son[j]==got[x]) continue;
             main(son[j],);
         }
         ),skp=got[x];
         calc(x,),skp=;
         ,c; i<a[x].size(); ++i) {
             ) ans[a[x][i].first]=;
             else ans[a[x][i].first]=ase[dep[x]+a[x][i].second].size();
         }
         );
     }
 }
 int main () {
     ; ];
     scanf("%d",&n);
     ; i<=n+; ++i) {
         scanf("%s%d",s,&x),++x;
         if (!rel[s]) rel[s]=++k;
         co[i]=rel[s];
         add(x,i);
     }
     scanf("%d",&q);
     ; i<=q; ++i) {
         scanf("%d%d",&x,&y),++x;
         a[x].push_back(make_pair(i,y));
     }
     dep[]=,siz[]=;
     tcd::dfs_ppw(,);
     tcd::main(,);
     ; i<=q; ++i) {
         printf("%d\n",ans[i]);
     }
     ;
 }