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 mxn = 2e3 + 5;
const long long mod = 1e9 + 7;
int Set(int N, int pos) { return N = N | (1 << pos); }
int reset(int N, int pos) { return N = N & ~(1 << pos); }
bool check(int N, int pos) { return (bool)(N & (1 << pos)); }
int val[mxn];
long long dp[mxn];
vector<int> adj[mxn];
int d;
void dfs(int u, int p, int x, int root) {
  dp[u] = 1;
  for (auto v : adj[u]) {
    if (v == p || abs(val[v] - x) > d || val[v] < x) continue;
    if (val[v] == x && v > root) continue;
    dfs(v, u, x, root);
    dp[u] = (dp[u] * (dp[v] + 1)) % mod;
    while (dp[u] < 0) dp[u] += mod;
  }
}
int main() {
  int n;
  cin >> d >> n;
  for (int i = 1; i <= n; i++) cin >> val[i];
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    memset(dp, 0, sizeof dp);
    dfs(i, i, val[i], i);
    ans = (ans + dp[i] % mod);
    ans %= mod;
    while (ans < 0) ans += mod;
  }
  ans = ans % mod;
  cout << ans << '\n';
  return 0;
}
