#include <bits/stdc++.h>
using namespace std;
const int inf = 1000 * 1000 * 1000 + 7;
const int maxn = 2012;
vector<long long> vec[maxn];
long long n, d, a[maxn], ans, dp[maxn], h;
bool mark[maxn];
int dfs(int v, int node, int root) {
  mark[v] = true;
  long long ret = 1;
  for (auto &u : vec[v])
    if (!mark[u] && node - a[u] <= d &&
        (a[u] < node || (a[u] == node && root < u)))
      ret = (long long)ret * (dfs(u, node, root) + 1) % inf;
  return ret;
}
int main() {
  cin >> d >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0, c, b; i < n - 1; i++)
    cin >> c >> b, c--, b--, vec[c].push_back(b), vec[b].push_back(c);
  for (int i = 0; i < n; i++) {
    dp[i] = dfs(i, a[i], i);
    memset(mark, 0, sizeof mark);
    ans = (long long)(ans + dp[i]) % inf;
  }
  cout << ans;
}

### Prompt

Develop a solution in CPP to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int inf = 1000 * 1000 * 1000 + 7;
const int maxn = 2012;
vector<long long> vec[maxn];
long long n, d, a[maxn], ans, dp[maxn], h;
bool mark[maxn];
int dfs(int v, int node, int root) {
  mark[v] = true;
  long long ret = 1;
  for (auto &u : vec[v])
    if (!mark[u] && node - a[u] <= d &&
        (a[u] < node || (a[u] == node && root < u)))
      ret = (long long)ret * (dfs(u, node, root) + 1) % inf;
  return ret;
}
int main() {
  cin >> d >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0, c, b; i < n - 1; i++)
    cin >> c >> b, c--, b--, vec[c].push_back(b), vec[b].push_back(c);
  for (int i = 0; i < n; i++) {
    dp[i] = dfs(i, a[i], i);
    memset(mark, 0, sizeof mark);
    ans = (long long)(ans + dp[i]) % inf;
  }
  cout << ans;
}

```