codeforces 375D：Tree and Queries

Description

You have a rooted tree consisting of n vertices. Each vertex of the tree has some color. We will assume that the tree vertices are numbered by integers from 1 to n. Then we represent the color of vertex v as c_v. The tree root is a vertex with number 1.

In this problem you need to answer to m queries. Each query is described by two integers v_j, k_j. The answer to query v_j, k_j is the number of such colors of vertices x, that the subtree of vertex v_j contains at least k_j vertices of color x.

You can find the definition of a rooted tree by the following link: http://en.wikipedia.org/wiki/Tree_(graph_theory).

Input

The first line contains two integers n and m (2 ≤ n ≤ 10⁵; 1 ≤ m ≤ 10⁵). The next line contains a sequence of integers c₁, c₂, ..., c_n (1 ≤ c_i ≤ 10⁵). The next n - 1 lines contain the edges of the tree. The i-th line contains the numbers a_i, b_i (1 ≤ a_i, b_i ≤ n; a_i ≠ b_i) — the vertices connected by an edge of the tree.

Next m lines contain the queries. The j-th line contains two integers v_j, k_j (1 ≤ v_j ≤ n; 1 ≤ k_j ≤ 10⁵).

Output

Print m integers — the answers to the queries in the order the queries appear in the input.

Examples

Input

8 5
1 2 2 3 3 2 3 3
1 2
1 5
2 3
2 4
5 6
5 7
5 8
1 2
1 3
1 4
2 3
5 3

Output

2
2
1
0
1

Input

4 1
1 2 3 4
1 2
2 3
3 4
1 1

Output

4

正解：莫队算法
解题报告：
　　今天心血来潮要刷莫队算法的题。
　　明明是一棵树，居然可以用莫队？！好吧是我大惊小怪了，dfs一遍，搞出dfs序，然后直接转成序列问题，询问、颜色全部转成序列的模式。
　　一样的维护就可以了。不过需要注意，我们记录每种颜色的出现次数，和至少出现某种次数的颜色个数。开始有一点想不通怎么O（1）更新至少某种次数，感觉要gi，后来发现，反正莫队一定是每次修改一位，那么最大贡献一定是1，那么我们一路更改下去就可以了，不虚。
　　好神，%%%。

 //It is made by jump~
 #include <iostream>
 #include <cstdlib>
 #include <cstring>
 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #include <ctime>
 #include <vector>
 #include <queue>
 #include <map>
 #include <set>
 #ifdef WIN32
 #define OT "%I64d"
 #else
 #define OT "%lld"
 #endif
 using namespace std;
 typedef long long LL;
 const int MAXN = ;
 const int MAXM = ;
 int n,m;
 int ans[MAXM];
 int yan[MAXN],col[MAXN];
 int first[MAXN],to[MAXN*],next[MAXN*];
 int L[MAXN],R[MAXN];
 int ecnt;
 int cnt[MAXM];//记录每种颜色的个数
 int jilu[MAXM];//至少k个颜色的个数
 int nowl,nowr;

 struct wen{
     int l,r,id,k,belong;
 }q[MAXM];

 inline int getint()
 {
        int w=,q=;
        char c=getchar();
        while((c<'' || c>'') && c!='-') c=getchar();
        if (c=='-')  q=, c=getchar();
        while (c>='' && c<='') w=w*+c-'', c=getchar();
        return q ? -w : w;
 }

 inline void dfs(int x,int fa){
     L[x]=++ecnt; col[ecnt]=yan[x];
     for(int i=first[x];i;i=next[i]) {
     int v=to[i];
     if(v!=fa) dfs(v,x);
     }
     R[x]=ecnt;
 }

 inline bool cmp(wen p,wen pp){ if(p.belong==pp.belong) return p.r<pp.r; return p.belong<pp.belong; }

 inline void add(int x,int type){
     if(type==) {
     cnt[col[x]]++;
     jilu[cnt[col[x]]]++;
     }else{
     jilu[cnt[col[x]]]--;
     cnt[col[x]]--;
     }
 }

 inline void work(){
     n=getint(); m=getint();
     for(int i=;i<=n;i++) yan[i]=getint();
     int x,y;
     for(int i=;i<n;i++) {
     x=getint(); y=getint();
     next[++ecnt]=first[x]; to[ecnt]=y; first[x]=ecnt;
     next[++ecnt]=first[y]; to[ecnt]=x; first[y]=ecnt;
     }
     ecnt=;  dfs(,);
     int size=sqrt(n);
     for(int i=;i<=m;i++) {
     x=getint(); q[i].l=L[x]; q[i].r=R[x]; q[i].id=i; q[i].k=getint();
     q[i].belong=(q[i].l-)/size+;
     }
     sort(q+,q+m+,cmp);
     for(int i=q[].l;i<=q[].r;i++) {
     cnt[col[i]]++;
     jilu[cnt[col[i]]]++;//只需一路添加就可以了，想想就知道并没有问题
     }
     ans[q[].id]=jilu[q[].k];
     nowl=q[].l; nowr=q[].r;
     for(int i=;i<=m;i++) {
     while(nowl<q[i].l) add(nowl,-),nowl++;
     while(nowl>q[i].l) add(nowl-,),nowl--;
     while(nowr<q[i].r) add(nowr+,),nowr++;
     while(nowr>q[i].r) add(nowr,-),nowr--;
     ans[q[i].id]=jilu[q[i].k];
     }
     for(int i=;i<=m;i++) printf("%d\n",ans[i]);
 }

 int main()
 {
   work();
   return ;
 }