#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e3 + 1;
const double Pi = acos(-1.0);
const long long INF = 1e18;
const long long MOD = 1e9 + 7;
int d, n;
int a[maxn];
long long dp[maxn];
int to[maxn], nxt[maxn], head[maxn];
int cnt = 0;
int now;
void add(int u, int v) {
  to[cnt] = v;
  nxt[cnt] = head[u];
  head[u] = cnt++;
}
void dfs(int x, int frt) {
  dp[x] = 1;
  for (int i = head[x]; i != -1; i = nxt[i])
    if (to[i] != frt) {
      if (a[now] - a[to[i]] < 0 || a[now] - a[to[i]] > d ||
          (a[now] - a[to[i]] == 0 && to[i] > now))
        continue;
      dfs(to[i], x);
      dp[x] = dp[x] * (dp[to[i]] + 1) % MOD;
    }
}
int main() {
  scanf("%d %d", &d, &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  memset(head, -1, sizeof(head));
  for (int i = 1; i < n; ++i) {
    int x, y;
    scanf("%d %d", &x, &y);
    add(x, y);
    add(y, x);
  }
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    now = i;
    dfs(i, -1);
    ans = (ans + dp[i]) % MOD;
  }
  printf("%lld\n", ans);
}

### Prompt

Your challenge is to write 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 maxn = 5e3 + 1;
const double Pi = acos(-1.0);
const long long INF = 1e18;
const long long MOD = 1e9 + 7;
int d, n;
int a[maxn];
long long dp[maxn];
int to[maxn], nxt[maxn], head[maxn];
int cnt = 0;
int now;
void add(int u, int v) {
  to[cnt] = v;
  nxt[cnt] = head[u];
  head[u] = cnt++;
}
void dfs(int x, int frt) {
  dp[x] = 1;
  for (int i = head[x]; i != -1; i = nxt[i])
    if (to[i] != frt) {
      if (a[now] - a[to[i]] < 0 || a[now] - a[to[i]] > d ||
          (a[now] - a[to[i]] == 0 && to[i] > now))
        continue;
      dfs(to[i], x);
      dp[x] = dp[x] * (dp[to[i]] + 1) % MOD;
    }
}
int main() {
  scanf("%d %d", &d, &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  memset(head, -1, sizeof(head));
  for (int i = 1; i < n; ++i) {
    int x, y;
    scanf("%d %d", &x, &y);
    add(x, y);
    add(y, x);
  }
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    now = i;
    dfs(i, -1);
    ans = (ans + dp[i]) % MOD;
  }
  printf("%lld\n", ans);
}

```