#include <bits/stdc++.h>
using namespace std;
const int maxn = 2020;
const int MOd = 1e9 + 7;
int a, d, dad[maxn], sz[maxn];
int ar[maxn];
vector<int> w[maxn];
long long dfs(int n, int back, int s) {
  long long ret = 1;
  int c = 0;
  for (int i = 0; i < w[n].size(); i++)
    if (w[n][i] != back && ((ar[w[n][i]] <= ar[s] + d && ar[s] < ar[w[n][i]]) ||
                            (ar[w[n][i]] == ar[s] && w[n][i] > s)))
      ret *= 1 + dfs(w[n][i], n, s), ret %= MOd, c = 1;
  return ret;
}
int main() {
  scanf("%d %d", &d, &a);
  for (int i = 1; i <= a; i++) scanf("%d", &ar[i]);
  for (int i = 1, j, k; i < a; i++)
    scanf("%d %d", &j, &k), w[j].push_back(k), w[k].push_back(j);
  long long ans = 0;
  for (int i = 1; i <= a; i++) ans += dfs(i, 0, i);
  cout << ans % MOd << endl;
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 maxn = 2020;
const int MOd = 1e9 + 7;
int a, d, dad[maxn], sz[maxn];
int ar[maxn];
vector<int> w[maxn];
long long dfs(int n, int back, int s) {
  long long ret = 1;
  int c = 0;
  for (int i = 0; i < w[n].size(); i++)
    if (w[n][i] != back && ((ar[w[n][i]] <= ar[s] + d && ar[s] < ar[w[n][i]]) ||
                            (ar[w[n][i]] == ar[s] && w[n][i] > s)))
      ret *= 1 + dfs(w[n][i], n, s), ret %= MOd, c = 1;
  return ret;
}
int main() {
  scanf("%d %d", &d, &a);
  for (int i = 1; i <= a; i++) scanf("%d", &ar[i]);
  for (int i = 1, j, k; i < a; i++)
    scanf("%d %d", &j, &k), w[j].push_back(k), w[k].push_back(j);
  long long ans = 0;
  for (int i = 1; i <= a; i++) ans += dfs(i, 0, i);
  cout << ans % MOd << endl;
  return 0;
}

```