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;
int const maxn = 2005;
int const mod = 1000000007;
int arr[maxn];
vector<int> adj[maxn];
int dp[maxn];
int ans = 0, d;
void dfs(int index, int fa, int mmax, int node) {
  for (int i = 0; i < adj[index].size(); i++) {
    int to = adj[index][i];
    if (to != fa && (mmax > arr[to] || (mmax == arr[to] && node < to)) &&
        (mmax - arr[to]) <= d) {
      dp[to] = dp[index];
      dfs(to, index, mmax, node);
      dp[index] += dp[to];
      dp[index] %= mod;
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, a, b;
  cin >> d >> n;
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
  }
  for (int i = 1; i < n; i++) {
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  for (int i = 1; i <= n; i++) {
    memset(dp, 0, sizeof(dp));
    dp[i] = 1;
    dfs(i, -1, arr[i], i);
    ans += dp[i];
    ans %= mod;
  }
  cout << ans << '\n';
  return 0;
}
