#include <bits/stdc++.h>
using namespace std;
const int maxn = 2500;
const int mod = (int)1e9 + 7;
int size[maxn], a[maxn], m[maxn], root;
vector<int> g[maxn];
bool was[maxn][maxn];
long long ans = 0;
void dfs(int v, int p = -1) {
  for (int u : g[v]) {
    if (u == p || !m[u]) continue;
    dfs(u, v);
  }
  long long res = 1;
  for (int u : g[v]) {
    if (m[u] && u != p) res = (res * 1ll * (size[u] + 1)) % mod;
  }
  size[v] = res;
}
int main() {
  int d, n, u, v;
  scanf("%d%d", &d, &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 1; i <= n; i++) {
    memset(size, 0, sizeof(size));
    memset(m, 0, sizeof(m));
    for (int j = 1; j <= n; j++)
      if ((a[i] < a[j] || (a[i] == a[j] && i < j)) && a[j] <= d + a[i])
        m[j] = 1;
    dfs(i);
    ans = (ans + size[i]) % mod;
  }
  printf("%d", ans);
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 maxn = 2500;
const int mod = (int)1e9 + 7;
int size[maxn], a[maxn], m[maxn], root;
vector<int> g[maxn];
bool was[maxn][maxn];
long long ans = 0;
void dfs(int v, int p = -1) {
  for (int u : g[v]) {
    if (u == p || !m[u]) continue;
    dfs(u, v);
  }
  long long res = 1;
  for (int u : g[v]) {
    if (m[u] && u != p) res = (res * 1ll * (size[u] + 1)) % mod;
  }
  size[v] = res;
}
int main() {
  int d, n, u, v;
  scanf("%d%d", &d, &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 1; i <= n; i++) {
    memset(size, 0, sizeof(size));
    memset(m, 0, sizeof(m));
    for (int j = 1; j <= n; j++)
      if ((a[i] < a[j] || (a[i] == a[j] && i < j)) && a[j] <= d + a[i])
        m[j] = 1;
    dfs(i);
    ans = (ans + size[i]) % mod;
  }
  printf("%d", ans);
  return 0;
}

```