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 Ada {
    static ArrayList<Integer> adj[];
    static boolean vis[];
    static long f[];
    static int a[];
    static int d,mod=(int)1e9 + 7;
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        d=sc.nextInt();int n= sc.nextInt();
        a=new int[n];
        for (int i=0;i<n;i++)a[i]=sc.nextInt();
        adj=new ArrayList[n];
        for (int i=0;i<n;i++)adj[i]=new ArrayList<>();
        for (int i=0;i<n-1;i++){
            int u=sc.nextInt()-1,v=sc.nextInt()-1;
            adj[u].add(v);
            adj[v].add(u);
        }
        long sum=0;
        vis=new boolean[n];
        f=new long[n];
        for (int i=0;i<n;i++){
            Arrays.fill(vis,false);
            Arrays.fill(f,0);
            dfs(i,i);
            sum+=(f[i]%mod);
        }
        System.out.println(sum%mod);
    }
    static void dfs(int v,int root){
        vis[v]=true;
        f[v]=1;
        for (int u:adj[v]){
            if (vis[u])continue;
            if (a[u]<a[root] || a[u]>a[root]+d)continue;
            if (a[u]==a[root] && (u<root))continue;
            dfs(u,root);
            f[v]=(f[v]*(f[u]+1))%mod;
        }
    }
}
