#include <bits/stdc++.h>
using namespace std;
const long long maxm = 2e3 + 5;
const long long mod = 1e9 + 7;
vector<long long> g[maxm];
long long a[maxm];
long long D, n;
long long d[maxm];
long long L, R;
long long rt;
void dfs(long long x, long long fa) {
  d[x] = 1;
  for (long long v : g[x]) {
    if (v == fa || a[v] < L || a[v] > R) continue;
    if (a[v] == a[rt]) {
      if (v > rt) continue;
    }
    dfs(v, x);
    d[x] = d[x] * (d[v] + 1) % mod;
  }
}
signed main() {
  ios::sync_with_stdio(0);
  cin >> D >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (long long i = 1; i < n; i++) {
    long long a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    L = a[i], R = a[i] + D;
    rt = i;
    dfs(i, i);
    ans = (ans + d[i]) % mod;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please formulate a CPP solution to the following problem:

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 long long maxm = 2e3 + 5;
const long long mod = 1e9 + 7;
vector<long long> g[maxm];
long long a[maxm];
long long D, n;
long long d[maxm];
long long L, R;
long long rt;
void dfs(long long x, long long fa) {
  d[x] = 1;
  for (long long v : g[x]) {
    if (v == fa || a[v] < L || a[v] > R) continue;
    if (a[v] == a[rt]) {
      if (v > rt) continue;
    }
    dfs(v, x);
    d[x] = d[x] * (d[v] + 1) % mod;
  }
}
signed main() {
  ios::sync_with_stdio(0);
  cin >> D >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (long long i = 1; i < n; i++) {
    long long a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    L = a[i], R = a[i] + D;
    rt = i;
    dfs(i, i);
    ans = (ans + d[i]) % mod;
  }
  cout << ans << endl;
  return 0;
}

```