#include <bits/stdc++.h>
using namespace std;
int n, d, a[2007];
const long long mod = 1e9 + 7;
vector<int> e[2007];
long long dp[2007];
bool used[2007];
void Clear() {
  for (int i = 0; i < 2007; i++) e[i].clear();
}
void add(int u, int v) {
  e[u].push_back(v);
  e[v].push_back(u);
}
void dfs(int u, int p, int x) {
  dp[u] = 1;
  for (int i = 0; i < e[u].size(); i++) {
    int v = e[u][i];
    if (v == p) continue;
    if (a[v] > a[x]) continue;
    if (a[v] < max(0, a[x] - d)) continue;
    if (a[v] == a[x] && v < x) continue;
    dfs(v, u, x);
    dp[u] *= dp[v] + 1LL;
    dp[u] %= mod;
  }
}
int main() {
  while (~scanf("%d%d", &d, &n)) {
    Clear();
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 1; i < n; i++) {
      int x, y;
      scanf("%d%d", &x, &y);
      add(x, y);
    }
    long long ans = 0;
    memset(used, 0, sizeof(used));
    for (int i = 1; i <= n; i++) {
      dfs(i, -1, i);
      ans += dp[i];
      ans %= mod;
    }
    printf("%lld\n", ans);
  }
}

### 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;
int n, d, a[2007];
const long long mod = 1e9 + 7;
vector<int> e[2007];
long long dp[2007];
bool used[2007];
void Clear() {
  for (int i = 0; i < 2007; i++) e[i].clear();
}
void add(int u, int v) {
  e[u].push_back(v);
  e[v].push_back(u);
}
void dfs(int u, int p, int x) {
  dp[u] = 1;
  for (int i = 0; i < e[u].size(); i++) {
    int v = e[u][i];
    if (v == p) continue;
    if (a[v] > a[x]) continue;
    if (a[v] < max(0, a[x] - d)) continue;
    if (a[v] == a[x] && v < x) continue;
    dfs(v, u, x);
    dp[u] *= dp[v] + 1LL;
    dp[u] %= mod;
  }
}
int main() {
  while (~scanf("%d%d", &d, &n)) {
    Clear();
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 1; i < n; i++) {
      int x, y;
      scanf("%d%d", &x, &y);
      add(x, y);
    }
    long long ans = 0;
    memset(used, 0, sizeof(used));
    for (int i = 1; i <= n; i++) {
      dfs(i, -1, i);
      ans += dp[i];
      ans %= mod;
    }
    printf("%lld\n", ans);
  }
}

```