#include <bits/stdc++.h>
using namespace std;
const int N = 2005;
int n, d, x, y, i, j, ans, cur, a[N];
int f[N], minim, maxim;
pair<int, int> b[N];
vector<int> g[N];
int go(int u, int p) {
  if (a[u] > maxim || a[u] < minim || f[u]) return 1;
  int e, i, to;
  long long s = 1;
  for (i = 0; i < g[u].size(); i++) {
    to = g[u][i];
    if (to == p) continue;
    e = go(to, u);
    s = (s * e) % ((long long)1000000007);
  }
  if (p == -1) return (int)s;
  return ((int)s + 1) % 1000000007;
}
int main() {
  scanf("%d%d", &d, &n);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  for (i = 1; i <= n; i++) b[i] = make_pair(a[i], i);
  sort(b + 1, b + n + 1);
  for (i = 1; i <= n; i++) {
    minim = a[b[i].second];
    maxim = a[b[i].second] + d;
    cur = go(b[i].second, -1);
    f[b[i].second] = 1;
    ans = (ans + cur) % 1000000007;
  }
  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 int N = 2005;
int n, d, x, y, i, j, ans, cur, a[N];
int f[N], minim, maxim;
pair<int, int> b[N];
vector<int> g[N];
int go(int u, int p) {
  if (a[u] > maxim || a[u] < minim || f[u]) return 1;
  int e, i, to;
  long long s = 1;
  for (i = 0; i < g[u].size(); i++) {
    to = g[u][i];
    if (to == p) continue;
    e = go(to, u);
    s = (s * e) % ((long long)1000000007);
  }
  if (p == -1) return (int)s;
  return ((int)s + 1) % 1000000007;
}
int main() {
  scanf("%d%d", &d, &n);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  for (i = 1; i <= n; i++) b[i] = make_pair(a[i], i);
  sort(b + 1, b + n + 1);
  for (i = 1; i <= n; i++) {
    minim = a[b[i].second];
    maxim = a[b[i].second] + d;
    cur = go(b[i].second, -1);
    f[b[i].second] = 1;
    ans = (ans + cur) % 1000000007;
  }
  cout << ans << endl;
  return 0;
}

```