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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 - 1lines 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

题意：
给定一颗树，树上的每个节点都有一个颜色，以1为根节点。
每次询问，问以节点v为根节点的子树里面，有多少种颜色出现次数大于k。
思路：
想到莫队之后，dfs序和树状数组很好想了。

#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#define fuck(x) cout<<#x<<" = "<<x<<endl;
#define debug(a,i) cout<<#a<<"["<<i<<"] = "<<a[i]<<endl;
#define ls (t<<1)
#define rs ((t<<1)+1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 100086;
const int maxm = 100086;
const int inf = 2.1e9;
const ll Inf = 999999999999999999;
const int mod = 1000000007;
const double eps = 1e-6;
const double pi = acos(-1);

int col[maxn];
int num[maxn];
int vis[maxn];
int cnt,Head[maxn];
int tl[maxn],tr[maxn];
struct node{
    int v,nxt;
}e[2*maxn];
struct query{
    int l,r,id,k;
}a[maxn];

void add_edge(int u,int v){
    e[cnt].v=v;
    e[cnt].nxt = Head[u];
    Head[u]=cnt++;
}

void dfs(int u,int fa){
    tl[u]=++cnt;
    num[cnt]=col[u];
    for(int k=Head[u];~k;k=e[k].nxt){
        if(e[k].v!=fa)dfs(e[k].v,u);
    }
    tr[u]=cnt;
}
int block;
bool cmp(query a,query b){
    if(a.l/block!=b.l/block){return a.l<b.l;}
    return a.r<b.r;
}
int lowbit(int x){
    return x&-x;
}

int bit[maxn];

int query(int pos){
    int ans=0;
    while(pos){
        ans+=bit[pos];
        pos-=lowbit(pos);
    }
    return ans;
}

void update(int pos,int val){
    if(pos<=0){return;}
    while(pos<maxn){
        bit[pos]+=val;
        pos+=lowbit(pos);
    }
}

int ans[maxn];

int main()
{
    memset(Head,-1,sizeof(Head));
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++){
        scanf("%d",&col[i]);
    }

    for(int i=1;i<n;i++){
        int x,y;
        scanf("%d%d",&x,&y);
        add_edge(x,y);
        add_edge(y,x);
    }
    cnt=0;
    dfs(1,0);

    block=sqrt(cnt);
    for(int i=1;i<=m;i++){
        int x,y;
        scanf("%d%d",&x,&y);
        a[i].l=tl[x];
        a[i].r=tr[x];
        a[i].k=y;
        a[i].id=i;
    }

    sort(a+1,a+1+m,cmp);
    int L=1,R=0;
    for(int i=1;i<=m;i++){
        while(L<a[i].l){
            update(vis[num[L]],-1);
            vis[num[L]]--;
            update(vis[num[L]],1);
            L++;
        }
        while(R<a[i].r){
            R++;
            update(vis[num[R]],-1);
            vis[num[R]]++;
            update(vis[num[R]],1);
        }
        while(L>a[i].l){
            L--;
            update(vis[num[L]],-1);
            vis[num[L]]++;
            update(vis[num[L]],1);
        }
        while(R>a[i].r){
            update(vis[num[R]],-1);
            vis[num[R]]--;
            update(vis[num[R]],1);
            R--;
        }
        ans[a[i].id]=query(maxn-1)-query(a[i].k-1);
    }
    for(int i=1;i<=m;i++){
        printf("%d
",ans[i]);
    }
    return 0;
}