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 int MOD = 1e9 + 7;
const int MAXN = 2e3 + 5;
long long dp[MAXN];
vector<int> adj[MAXN];
int ar[MAXN];
int src;
int d;
bool done[MAXN][MAXN];
void DFS(int s, int par = -1) {
  dp[s] = 1;
  for (auto it : adj[s]) {
    if (it == par) continue;
    if (ar[it] < ar[src]) continue;
    if (ar[it] - ar[src] > d) continue;
    if (done[it][src]) continue;
    done[it][src] = done[src][it] = true;
    DFS(it, s);
    dp[s] *= 1LL + dp[it];
    if (dp[s] >= MOD) dp[s] %= MOD;
  }
}
int main() {
  int n;
  cin >> d >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &ar[i]);
  }
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    src = i;
    DFS(i);
    ans += dp[i];
    if (ans >= MOD) ans %= MOD;
  }
  cout << ans << endl;
  return 0;
}
