#include <bits/stdc++.h>
using namespace std;
const int MAX = 2014;
vector<int> gr[MAX];
int a[MAX];
bitset<MAX> viz;
const int MOD = 1000000007;
inline long long dfs(int nod, int maxim, int boss, int d) {
  long long S = 1;
  viz[nod] = 1;
  for (auto x : gr[nod])
    if (!viz[x] and a[x] >= maxim - d and
        (a[x] < maxim or (a[x] == maxim and x < boss)))
      S = 1LL * S * (dfs(x, maxim, boss, d) + 1) % MOD;
  viz[nod] = 0;
  return S;
}
int main() {
  ios ::sync_with_stdio(false);
  int d, n;
  cin >> d >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  int m = n - 1;
  while (m--) {
    int x, y;
    cin >> x >> y;
    gr[x].push_back(y);
    gr[y].push_back(x);
  }
  long long CET = 0;
  for (int i = 1; i <= n; ++i)
    CET = (1LL * CET + dfs(i, a[i], i, d) % MOD) % MOD;
  cout << CET;
  return 0;
}

### Prompt

In cpp, your task is to solve 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 int MAX = 2014;
vector<int> gr[MAX];
int a[MAX];
bitset<MAX> viz;
const int MOD = 1000000007;
inline long long dfs(int nod, int maxim, int boss, int d) {
  long long S = 1;
  viz[nod] = 1;
  for (auto x : gr[nod])
    if (!viz[x] and a[x] >= maxim - d and
        (a[x] < maxim or (a[x] == maxim and x < boss)))
      S = 1LL * S * (dfs(x, maxim, boss, d) + 1) % MOD;
  viz[nod] = 0;
  return S;
}
int main() {
  ios ::sync_with_stdio(false);
  int d, n;
  cin >> d >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  int m = n - 1;
  while (m--) {
    int x, y;
    cin >> x >> y;
    gr[x].push_back(y);
    gr[y].push_back(x);
  }
  long long CET = 0;
  for (int i = 1; i <= n; ++i)
    CET = (1LL * CET + dfs(i, a[i], i, d) % MOD) % MOD;
  cout << CET;
  return 0;
}

```