#include <bits/stdc++.h>
using namespace std;
int n;
vector<int> adj[2123];
long long a[2123], t[2123], d;
int vis[2123], curr;
void dfs(int u, int p) {
  t[u] = 1LL;
  for (int v : adj[u]) {
    if ((v != p) && (a[curr] <= a[v]) && (a[v] <= (a[curr] + d))) {
      if (v < curr && a[curr] == a[v]) continue;
      dfs(v, u);
      t[u] = (t[u] * (t[v] + 1LL)) % 1000000007LL;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  int u, v;
  cin >> d >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i < n; ++i) {
    cin >> u >> v;
    adj[u].emplace_back(v);
    adj[v].emplace_back(u);
  }
  long long ans = 0LL;
  for (int i = 1; i <= n; ++i) {
    curr = i;
    for (int j = 1; j <= n; ++j) t[j] = 0;
    dfs(i, -1);
    ans += t[i];
    if (ans >= 1000000007LL) ans -= 1000000007LL;
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Your challenge is to write 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;
int n;
vector<int> adj[2123];
long long a[2123], t[2123], d;
int vis[2123], curr;
void dfs(int u, int p) {
  t[u] = 1LL;
  for (int v : adj[u]) {
    if ((v != p) && (a[curr] <= a[v]) && (a[v] <= (a[curr] + d))) {
      if (v < curr && a[curr] == a[v]) continue;
      dfs(v, u);
      t[u] = (t[u] * (t[v] + 1LL)) % 1000000007LL;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  int u, v;
  cin >> d >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i < n; ++i) {
    cin >> u >> v;
    adj[u].emplace_back(v);
    adj[v].emplace_back(u);
  }
  long long ans = 0LL;
  for (int i = 1; i <= n; ++i) {
    curr = i;
    for (int j = 1; j <= n; ++j) t[j] = 0;
    dfs(i, -1);
    ans += t[i];
    if (ans >= 1000000007LL) ans -= 1000000007LL;
  }
  cout << ans << '\n';
  return 0;
}

```