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;
vector<int> V[2222];
int val[2222];
long long dp[2222];
int d;
bool vis[2222];
void dfs(int u, int fa) {
  int sz = V[u].size();
  vis[u] = true;
  for (int i = 0; i < sz; i++) {
    int v = V[u][i];
    if (vis[v]) continue;
    if ((val[v] > val[fa] && val[v] <= val[fa] + d) ||
        (val[v] == val[fa] && v > fa)) {
      dfs(v, fa);
      dp[u] = (dp[u] * (dp[v] + 1)) % 1000000007;
    }
  }
}
int main() {
  int n;
  while (cin >> d >> n) {
    for (int i = 1; i <= n; i++) V[i].clear();
    for (int i = 1; i <= n; i++) cin >> val[i];
    int a, b;
    for (int i = 0; i < n - 1; i++) {
      cin >> a >> b;
      V[a].push_back(b);
      V[b].push_back(a);
    }
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      memset(vis, false, sizeof(vis));
      for (int j = 1; j <= n; j++) dp[j] = 1;
      dfs(i, i);
      ans = (ans + dp[i]) % 1000000007;
    }
    cout << ans << endl;
  }
  return 0;
}
