#include <bits/stdc++.h>
using namespace std;
vector<int> Graph[2005];
int ranks[2005];
long long int dfs(int node, int low, int high, int parent = -1) {
  if (ranks[node] > high || ranks[node] < low) return 0;
  int i;
  long long int ans = 1;
  for (i = 0; i < Graph[node].size(); i++) {
    if (Graph[node][i] == parent) continue;
    ans = (ans * (1 + dfs(Graph[node][i], low, high, node))) % 1000000007;
  }
  return ans;
}
int main() {
  int n, i, j, d;
  scanf("%d %d", &d, &n);
  pair<int, int> p[n];
  for (i = 0; i < n; i++) {
    p[i].second = i + 1;
    scanf("%d", &p[i].first);
  }
  sort(p, p + n);
  for (i = 0; i < n; i++) {
    ranks[p[i].second] = i;
  }
  for (i = 1; i < n; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    Graph[u].push_back(v);
    Graph[v].push_back(u);
  }
  long long int ans = 0;
  for (i = 0, j = 0; i < n; i++) {
    for (; j < n && p[j].first - p[i].first <= d; j++)
      ;
    ans = (ans + dfs(p[i].second, i, j - 1)) % 1000000007;
  }
  printf("%I64d", ans);
  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;
vector<int> Graph[2005];
int ranks[2005];
long long int dfs(int node, int low, int high, int parent = -1) {
  if (ranks[node] > high || ranks[node] < low) return 0;
  int i;
  long long int ans = 1;
  for (i = 0; i < Graph[node].size(); i++) {
    if (Graph[node][i] == parent) continue;
    ans = (ans * (1 + dfs(Graph[node][i], low, high, node))) % 1000000007;
  }
  return ans;
}
int main() {
  int n, i, j, d;
  scanf("%d %d", &d, &n);
  pair<int, int> p[n];
  for (i = 0; i < n; i++) {
    p[i].second = i + 1;
    scanf("%d", &p[i].first);
  }
  sort(p, p + n);
  for (i = 0; i < n; i++) {
    ranks[p[i].second] = i;
  }
  for (i = 1; i < n; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    Graph[u].push_back(v);
    Graph[v].push_back(u);
  }
  long long int ans = 0;
  for (i = 0, j = 0; i < n; i++) {
    for (; j < n && p[j].first - p[i].first <= d; j++)
      ;
    ans = (ans + dfs(p[i].second, i, j - 1)) % 1000000007;
  }
  printf("%I64d", ans);
  return 0;
}

```