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.*;

import static java.lang.Math.*;

public class Main extends PrintWriter {
    BufferedReader in;
    StringTokenizer stok;
    final Random rand = new Random(31);
    final int inf = (int) 1e9;
    final long linf = (long) 1e18;
    final int mod = inf + 7;

    List<Integer>[] g;
    boolean[] used;
    int[] a;

    int dfs(int v, int p, int l, int r) {
        long res = 1;
        for (int i : g[v]) {
            if (i != p && !used[i] && l <= a[i] && a[i] <= r) {
                res *= (dfs(i, v, l, r) + 1);
                res %= mod;
            }
        }
        return (int) res;
    }

    public void solve() throws IOException {
        int d = nextInt();
        int n = nextInt();
        a = nextIntArray(n);
        g = createAdjacencyList(n);
        for (int i = 0; i < n - 1; i++) {
            int x = nextInt() - 1;
            int y = nextInt() - 1;
            g[x].add(y);
            g[y].add(x);
        }
        used = new boolean[n];
        int ans = 0;
        for (int min = 0; min <= 2000; min++) {
            Arrays.fill(used, false);
            for (int i = 0; i < n; i++) {
                if (a[i] == min) {
                    used[i] = true;
                    ans += dfs(i, -1, min - d, min);
                    ans %= mod;
                }
            }
        }
        println(ans);
    }

    public void run() {
        try {
            solve();
            close();
        } catch (Exception e) {
            e.printStackTrace();
            System.exit(abs(-1));
        }
    }

    Main() throws IOException {
        super(System.out);
        in = new BufferedReader(new InputStreamReader(System.in));
    }

    Main(String s) throws IOException {
        super("".equals(s) ? "output.txt" : (s + ".out"));
        in = new BufferedReader(new FileReader("".equals(s) ? "input.txt" : (s + ".in")));
    }

    public static void main(String[] args) throws IOException {
        try {
            Locale.setDefault(Locale.US);
        } catch (Exception ignored) {
        }
        new Main().run();
    }

    String next() throws IOException {
        while (stok == null || !stok.hasMoreTokens()) {
            String s = in.readLine();
            if (s == null) {
                return null;
            }
            stok = new StringTokenizer(s);
        }
        return stok.nextToken();
    }

    int nextInt() throws IOException {
        return Integer.parseInt(next());
    }

    long nextLong() throws IOException {
        return Long.parseLong(next());
    }

    double nextDouble() throws IOException {
        return Double.parseDouble(next());
    }

    int[] nextIntArray(int len) throws IOException {
        int[] a = new int[len];
        for (int i = 0; i < len; i++) {
            a[i] = nextInt();
        }
        return a;
    }

    void shuffle(int[] a) {
        for (int i = 1; i < a.length; i++) {
            int x = rand.nextInt(i + 1);
            int t = a[i];
            a[i] = a[x];
            a[x] = t;
        }
    }

    boolean nextPerm(int[] p) {
        for (int a = p.length - 2; a >= 0; --a)
            if (p[a] < p[a + 1])
                for (int b = p.length - 1; ; --b)
                    if (p[b] > p[a]) {
                        int t = p[a];
                        p[a] = p[b];
                        p[b] = t;
                        for (++a, b = p.length - 1; a < b; ++a, --b) {
                            t = p[a];
                            p[a] = p[b];
                            p[b] = t;
                        }
                        return true;
                    }
        return false;
    }

    <T> List<T>[] createAdjacencyList(int countVertex) {
        List<T>[] res = new List[countVertex];
        for (int i = 0; i < countVertex; i++) {
            res[i] = new ArrayList<T>();
        }
        return res;
    }
}
