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.BufferedInputStream;
import java.util.Scanner;

/**
 * Created by liverliu on 14/11/17.
 */
public class Main {

    static class Node {
        int val;
    }
    static class Edge {
        public int u, v, next;
    }
    static Node node[] = new Node[2005];
    static Edge edge[] = new Edge[4010];
    static int head[] = new int[2005], E;
    static void addEdge(int u, int v) {
        edge[E].u = u;
        edge[E].v = v;
        edge[E].next = head[u];
        head[u] = E++;
    }

    static final int mod = 1000000007;
    static int d,n;

    static void init() {
        E = 0;
        for(int i=0;i<n;i++) {
            head[i] = -1;
            node[i] = new Node();
            edge[i*2] = new Edge();
            edge[i*2+1] = new Edge();
        }
    }

    static long dfs(int u, int fa, int rt) {
        long ans = 1;
        for(int i=head[u]; i!=-1; i=edge[i].next) {
            int v = edge[i].v;
            if(fa == v || node[v].val > node[rt].val || (node[v].val == node[rt].val && v > rt) ||
                    node[rt].val-node[v].val > d) {
                continue;
            }
            long tmp = dfs(v, u, rt);
            ans = (ans * (tmp + 1)) %mod;
        }
        return ans;
    }

    public static void main(String args[]) {

        Scanner scan = new Scanner(new BufferedInputStream(System.in));
        while(scan.hasNext()) {
            d = scan.nextInt();
            n = scan.nextInt();
            init();
            for(int i=0;i<n;i++) {
                node[i].val = scan.nextInt();
            }
            for(int i=0;i<n-1;i++) {
                int u = scan.nextInt();
                int v = scan.nextInt();
                addEdge(--u, --v);
                addEdge(v, u);
            }
            long ans=0;
            for(int i=0;i<n;i++) {
                ans += dfs(i, -1, i);
                ans %= mod;
            }
            System.out.println(ans);
        }
    }
}
