Problem: 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 cv. 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 vj, kj. The answer to query vj, kj is the number of such colors of vertices x, that the subtree of vertex vj contains at least kj 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, ..., cn (1 ≤ ci ≤ 105). The next n - 1 lines contain the edges of the tree. The i-th line contains the numbers ai, bi (1 ≤ ai, bi ≤ n; ai ≠ bi) — the vertices connected by an edge of the tree.

Next m lines contain the queries. The j-th line contains two integers vj, kj (1 ≤ vj ≤ n; 1 ≤ kj ≤ 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

Note

A subtree of vertex v in a rooted tree with root r is a set of vertices {u : dist(r, v) + dist(v, u) = dist(r, u)}. Where dist(x, y) is the length (in edges) of the shortest path between vertices x and y.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
public class Main
{
	int n,m,c[],start[],end[];
	ArrayList<Integer> list[],euler;
	private void solve()throws Exception
	{
		n=nextInt();
		m=nextInt();
		c=new int[n+1];
		for(int i=1;i<=n;i++)
			c[i]=nextInt();
		list=new ArrayList[n+1];
		for(int i=1;i<=n;i++)
			list[i]=new ArrayList<>();
		for(int i=1;i<n;i++)
		{
			int a=nextInt();
			int b=nextInt();
			list[a].add(b);
			list[b].add(a);
		}
		euler=new ArrayList<>();
		euler.add(0);
		start=new int[n+1];
		end=new int[n+1];
		dfs(1,0);
		n=euler.size()-1;
		Query queries[]=new Query[m+1];
		for(int i=1;i<=m;i++)
		{
			int v=nextInt();
			int k=nextInt();
			queries[i]=new Query(start[v],end[v],k,i);
		}
		MoAlgorithm mo=new MoAlgorithm(euler,c,queries);
		int ans[]=mo.solve();
		for(int i=1;i<=m;i++)
			out.println(ans[i]);
	}
	void dfs(int v,int p)
	{
		start[v]=euler.size();
		euler.add(v);
		for(int vv:list[v])
			if(vv!=p)
				dfs(vv,v);
		end[v]=euler.size()-1;
	}
	class MoAlgorithm
	{
		final int maxn=(int)(1e5);
		int cnt[],freq[],a[],BS,n,m,res[];
		Query queries[];
		ArrayList<Integer> euler;
		int c[];
		MoAlgorithm(ArrayList<Integer> euler,int c[],Query qu[])
		{
			this.euler=euler;
			this.c=c;
			queries=qu;
			n=euler.size()-1;
			m=queries.length-1;
			BS=(int)Math.ceil(Math.sqrt(n));
			Arrays.sort(queries,1,m+1,(p,q)->{
				int bp=(p.l+BS-1)/BS;
				int bq=(q.l+BS-1)/BS;
				if(bp==bq)
					return p.r-q.r;
				else
					return bp-bq;
			});
		}
		int[] solve()
		{
			res=new int[m+1];
			cnt=new int[maxn+1];
			freq=new int[maxn+1];
			int L=1,R=0;
			for(int i=1;i<=m;i++)
			{
				Query q=queries[i];
				while(R<q.r)
					add(c[euler.get(++R)]);
				while(L>q.l)
					add(c[euler.get(--L)]);
				while(L<q.l)
					remove(c[euler.get(L++)]);
				while(R>q.r)
					remove(c[euler.get(R--)]);
				res[q.idx]=freq[q.k];
			}
			return res;
		}
		void add(int val)
		{
			cnt[val]++;
			freq[cnt[val]]++;
		}
		void remove(int val)
		{
			freq[cnt[val]]--;
			cnt[val]--;
		}
	}
	class Query
	{
		int l,r,k,idx;
		Query(int l,int r,int k,int idx)
		{
			this.l=l;
			this.r=r;
			this.k=k;
			this.idx=idx;
		}
	}

	 
	///////////////////////////////////////////////////////////

	public void run()throws Exception
	{
		br=new BufferedReader(new InputStreamReader(System.in));
		st=null;
		out=new PrintWriter(System.out);

		solve();
		
		br.close();
		out.close();
	}
	public static void main(String args[])throws Exception{
		new Main().run();
	}
	BufferedReader br;
	StringTokenizer st;
	PrintWriter out;
	String nextToken()throws Exception{
		while(st==null || !st.hasMoreTokens())
		st=new StringTokenizer(br.readLine());
		return st.nextToken();
	}
	String nextLine()throws Exception{
		return br.readLine();
	}
	int nextInt()throws Exception{
		return Integer.parseInt(nextToken());
	}
	long nextLong()throws Exception{
		return Long.parseLong(nextToken());
	}
	double nextDouble()throws Exception{
		return Double.parseDouble(nextToken());
	}
}