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. .

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).

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.

Print the number of valid sets modulo 1000000007.

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.

题意：给你一个n点的树，和每个点的权值，问你多少种子树满足(最大权值点-最小权值点)<=d

题解：定义dp[i]表示以i为最小权值根节点的子树方案数，注意维护此条件

于是答案就是  ∑dp[i] %mod (1<=i<=n);

///

#include<bits/stdc++.h>

using namespace std ;

typedef long long ll;

#define mem(a) memset(a,0,sizeof(a))

#define pb push_back

#define meminf(a) memset(a,127,sizeof(a));

inline ll read()

{

    ll x=,f=;char ch=getchar();

    while(ch<''||ch>''){

        if(ch=='-')f=-;ch=getchar();

    }

    while(ch>=''&&ch<=''){

        x=x*+ch-'';ch=getchar();

    }return x*f;

}

//****************************************

#define maxn 2000+50

#define mod 1000000007

#define inf 1000000007

int d,n,a[maxn],vis[maxn];

vector<int >G[maxn];

ll dp[maxn];//以i为最小根节点，的方案数

void dfs(int x,int pre){

   dp[x]=;vis[x]=;

   for(int i=;i<G[x].size();i++){

      if(!vis[G[x][i]]){

            if(a[G[x][i]]<a[pre]||a[G[x][i]]>a[pre]+d)continue;

            if(a[G[x][i]]==a[pre]&&G[x][i]<pre)continue;

            dfs(G[x][i],pre);

            dp[x]=(dp[x]*(dp[G[x][i]]+))%mod;

      }

   }

}

int main(){

    d=read(),n=read();

    for(int i=;i<=n;i++){

        scanf("%d",&a[i]);

    }int u,v;

    for(int i=;i<n;i++){

        scanf("%d%d",&u,&v);

        G[u].pb(v);G[v].pb(u);

    }ll ans=;

    for(int i=;i<=n;i++){

        mem(dp);mem(vis);

        dfs(i,i);

        ans=(ans+dp[i])%mod;

    }

    cout<<ans<<endl;

  return ;

}