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<vector<long long int>> graph(2003);
vector<long long int> visited(2003), dp(2003), visited1(2003);
long long int d, n;
vector<long long int> a(2003);
long long int ans = 0;
void dfs(long long int u, long long int val) {
  visited[u] = 1;
  dp[u] = 1;
  for (long long int i = 0; i < graph[u].size(); i++) {
    if (a[graph[u][i]] > val && a[graph[u][i]] <= val + d) {
      if (!visited[graph[u][i]]) {
        dfs(graph[u][i], val);
        dp[u] = (dp[u] * (dp[graph[u][i]] + 1)) % 1000000007;
      }
    } else if (a[graph[u][i]] == val && !visited1[graph[u][i]]) {
      if (!visited[graph[u][i]]) {
        dfs(graph[u][i], val);
        dp[u] = (dp[u] * (dp[graph[u][i]] + 1)) % 1000000007;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> d >> n;
  for (long long int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (long long int i = 0; i < n - 1; i++) {
    long long int u, v;
    cin >> u >> v;
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  for (int i = 1; i <= n; i++) {
    visited1[i] = 1;
    fill(visited.begin(), visited.end(), 0);
    fill(dp.begin(), dp.end(), 0);
    dfs(i, a[i]);
    ans = (ans + dp[i]) % 1000000007;
  }
  cout << ans << "\n";
}
