As you know, an undirected connected graph with n nodes and n - 1 edges is called a tree. You are given an integer d and a tree consisting of n nodes. Each node i has a value ai associated with it.

We call a set S of tree nodes valid if following conditions are satisfied:

  1. S is non-empty.
  2. S is connected. In other words, if nodes u and v are in S, then all nodes lying on the simple path between u and v should also be presented in S.
  3. <image>.



Your task is to count the number of valid sets. Since the result can be very large, you must print its remainder modulo 1000000007 (109 + 7).

Input

The first line contains two space-separated integers d (0 ≤ d ≤ 2000) and n (1 ≤ n ≤ 2000).

The second line contains n space-separated positive integers a1, a2, ..., an(1 ≤ ai ≤ 2000).

Then the next n - 1 line each contain pair of integers u and v (1 ≤ u, v ≤ n) denoting that there is an edge between u and v. It is guaranteed that these edges form a tree.

Output

Print the number of valid sets modulo 1000000007.

Examples

Input

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


Output

8


Input

0 3
1 2 3
1 2
2 3


Output

3


Input

4 8
7 8 7 5 4 6 4 10
1 6
1 2
5 8
1 3
3 5
6 7
3 4


Output

41

Note

In the first sample, there are exactly 8 valid sets: {1}, {2}, {3}, {4}, {1, 2}, {1, 3}, {3, 4} and {1, 3, 4}. Set {1, 2, 3, 4} is not valid, because the third condition isn't satisfied. Set {1, 4} satisfies the third condition, but conflicts with the second condition.
import java.util.*;
public class ProD {
	static int d,n,ee,a,b;
	static long ans,mod=1000000007;
	static int[] aa=new int[2005];
	static int[] head=new int[2005];
	static int[] edge=new int[4005];
	static int[] next=new int[4005];
	static int[] mm=new int[2005];
	static long[] dp=new long[2005];
	static void add(int a,int b)
	{
		edge[ee]=b;next[ee]=head[a];
		head[a]=ee++;
	}
	static void dfs(int u,int v)
	{
		int s;dp[u]=1;
		for(int i=head[u];i!=-1;i=next[i])
		{
			s=edge[i];
			if(mm[s]==0||s==v) continue;
			dfs(s,u);
			dp[u]=dp[u]*(dp[s]+1)%mod;
		}
	}

	public static void main(String[] args) {
		Scanner in=new Scanner(System.in);
		d=in.nextInt();n=in.nextInt();
		for(int i=1;i<=n;i++)
			aa[i]=in.nextInt();
		Arrays.fill(head,-1);ee=0;
		for(int i=1;i<n;i++)
		{
			a=in.nextInt();b=in.nextInt();
			add(a,b);add(b,a);
		}
		for(int i=1;i<=n;i++)
		{
			Arrays.fill(mm,0);
			for(int j=1;j<=n;j++)
			{
				if(aa[i]>aa[j]&&aa[i]-aa[j]<=d) mm[j]=1;
				else if(i>=j&&aa[i]==aa[j]) mm[j]=1;
			}
			Arrays.fill(dp,0);
			dfs(i,0);
			ans=(ans+dp[i])%mod;
		}
		System.out.println(ans);
	}
}
