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.io.*;
import java.util.*;

public class Main {

	public static class FastReader {
		BufferedReader br;
		StringTokenizer root;
		
 
		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
 
		String next() {
			while (root == null || !root.hasMoreTokens()) {
				try {
					root = new StringTokenizer(br.readLine());
				} catch (Exception addd) {
					addd.printStackTrace();
				}
			}
			return root.nextToken();
		}
 
		int nextInt() {
			return Integer.parseInt(next());
		}
 
		double nextDouble() {
			return Double.parseDouble(next());
		}
 
		long nextLong() {
			return Long.parseLong(next());
		}
 
		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (Exception addd) {
				addd.printStackTrace();
			}
			return str;
		}
	}
	
	public static PrintWriter out = new PrintWriter (new BufferedOutputStream(System.out));
	
	public static FastReader sc = new FastReader();
 
	static int mod = (int) (1e9+7),MAX=(int) (3e5);
	static List<Integer>[] edges ;
	static boolean[][] vis;
	static int[] val;
	static int d;
	
	public static void main(String[] args) {
		d = sc.nextInt();
		int n = sc.nextInt();
		vis = new boolean[n+1][n+1];
		val = new int[n+1];
		for(int i=1;i<=n;++i) val[i] = sc.nextInt();
		edges = new ArrayList[n+1];
		for(int i=0;i<=n;++i) edges[i] = new ArrayList<>();
		
		for(int i=1;i<n;++i) {
			int u = sc.nextInt();
			int v = sc.nextInt();
			edges[u].add(v);
			edges[v].add(u);
		}
	    long ans = 0;
	    for(int i=1;i<=n;++i) {
	    	ans+=dfs(i,0,i);
	    	if(ans >= mod) ans-=mod;
	    }
	    out.print(ans);
		out.close();
	}

	private static long dfs(int v, int par,int min) {
		long ans = 1;
		for(int child : edges[v]) {
			if(child == par) continue;
			if(val[child] < val[min] || val[child] > val[min] + d) continue;
			if(vis[child][min]) continue; // if already created a set of child to root
				ans *= (dfs(child,v,min)+1L)%mod;
				ans%=mod;
				vis[min][child] = true; // mark true for the set of root to child
			}
		
		return ans;
	}

	
}
