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

public class Main {
    static long mod = (long)(1e9) + 7;
    
    int n, d, w;
    int[] val = new int[2005];
    int[] head = new int[2005];
    int[] treefrom = new int[2005*2];
    int[] treeto = new int[2005*2];
    int[] treenext = new int[2005*2];
    boolean[][] vis = new boolean [2005][2005];
    
    public long dfs(int root, int now, int fa) {
        int i, to;
        long ans = 1;
        for(i = head[now]; i != -1; i = treenext[i]) {
            to = treeto[i];
            if(to == fa) continue;
            if(Math.abs(val[root] - val[to]) > d) continue;
            if(val[to] > val[root]) continue;
            if(val[to] < val[root]) {
                ans *= dfs(root, to, now)+1;
                ans %= mod;
            }
            else if(val[to] == val[root]){
                if(vis[root][to] || vis[to][root]) continue;
                vis[root][to]=vis[to][root]=true;
                vis[root][to] = vis[to][root] = true;
                ans *= dfs(root,to,now)+1;
                ans %= mod;
            }
        }
        return ans;
    }
    public void work() {
        
        d = cin.nextInt();
        n = cin.nextInt();
        w = 0;
        for(int i = 1; i <= n; i ++) {
            val[i] = cin.nextInt();
        }
        for(int i = 1; i <= n; i ++) {
            head[i] = -1;
        }

        for(int i = 1, u, v; i < n; i ++) {
            u = cin.nextInt();
            v = cin.nextInt();
            w++;
            treefrom[w]=u;
            treeto[w]=v;
            treenext[w]=head[u];
            head[u]=w;
            
            w++;
            treefrom[w]=v;
            treeto[w]=u;
            treenext[w]=head[v];
            head[v]=w;
        }
        long ans = 0;
        for(int i = 1; i <= n; i ++) {
            ans += dfs(i, i, 0);
            ans %= mod;
        }
        System.out.println(ans);
    }

    Main() {
    //  out = new PrintWriter(System.out);
        cin = new Scanner(System.in);
    }

    public static void main(String[] args) {
        Main e = new Main();
        e.work();
    }

    public Scanner cin;
//  public PrintWriter out;
}
