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.
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1000000007;
int n, d, a[3000], used[3000], k;
long long f[3000];
vector<int> g[3000];
void init() {
  for (int i = 0; i < n; i++) {
    f[i] = 1;
    used[i] = 0;
  }
}
void dfs(int v) {
  used[v] = 1;
  for (int i = 0; i < g[v].size(); i++)
    if (used[g[v][i]] == 0 && a[k] <= a[g[v][i]] && a[g[v][i]] <= a[k] + d) {
      if (a[k] == a[g[v][i]] && k < g[v][i]) continue;
      dfs(g[v][i]);
      f[v] *= (f[g[v][i]] + 1);
      f[v] %= INF;
    }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> d >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    init();
    k = i;
    dfs(i);
    ans += f[i];
    ans %= INF;
  }
  cout << ans << endl;
  return 0;
}
