#include <bits/stdc++.h>
using namespace std;
const int MAX = 2000 + 5;
const int mod = 1e9 + 7;
int n, d;
long long ans;
long long val[MAX];
vector<int> p[MAX];
long long dfs(int x, int fa, int mx, int begin) {
  int i, v;
  long long cnt, sum;
  sum = 1;
  for (i = 0; i < p[x].size(); i++) {
    cnt = 0;
    v = p[x][i];
    if (v == fa) continue;
    if (val[v] > mx || mx - val[v] > d)
      continue;
    else if (val[v] == val[begin] && v < begin)
      continue;
    else {
      cnt = dfs(v, x, mx, begin);
      cnt++;
    }
    sum = (sum * cnt) % mod;
  }
  return sum;
}
int main() {
  int i;
  int u, v;
  scanf("%d%d", &d, &n);
  for (i = 1; i <= n; i++) scanf("%d", &val[i]);
  for (i = 0; i < n - 1; i++) {
    scanf("%d%d", &u, &v);
    p[u].push_back(v);
    p[v].push_back(u);
  }
  for (i = 1; i <= n; i++) {
    ans = (ans + dfs(i, 0, val[i], i)) % mod;
  }
  printf("%lld\n", ans);
  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 MAX = 2000 + 5;
const int mod = 1e9 + 7;
int n, d;
long long ans;
long long val[MAX];
vector<int> p[MAX];
long long dfs(int x, int fa, int mx, int begin) {
  int i, v;
  long long cnt, sum;
  sum = 1;
  for (i = 0; i < p[x].size(); i++) {
    cnt = 0;
    v = p[x][i];
    if (v == fa) continue;
    if (val[v] > mx || mx - val[v] > d)
      continue;
    else if (val[v] == val[begin] && v < begin)
      continue;
    else {
      cnt = dfs(v, x, mx, begin);
      cnt++;
    }
    sum = (sum * cnt) % mod;
  }
  return sum;
}
int main() {
  int i;
  int u, v;
  scanf("%d%d", &d, &n);
  for (i = 1; i <= n; i++) scanf("%d", &val[i]);
  for (i = 0; i < n - 1; i++) {
    scanf("%d%d", &u, &v);
    p[u].push_back(v);
    p[v].push_back(u);
  }
  for (i = 1; i <= n; i++) {
    ans = (ans + dfs(i, 0, val[i], i)) % mod;
  }
  printf("%lld\n", ans);
  return 0;
}

```