For a finite set of integers X, let f(X)=\max X - \min X.

Given are N integers A_1,...,A_N.

We will choose K of them and let S be the set of the integers chosen. If we distinguish elements with different indices even when their values are the same, there are {}_N C_K ways to make this choice. Find the sum of f(S) over all those ways.

Since the answer can be enormous, print it \bmod (10^9+7).

Constraints

* 1 \leq N \leq 10^5
* 1 \leq K \leq N
* |A_i| \leq 10^9

Input

Input is given from Standard Input in the following format:


N K
A_1 ... A_N


Output

Print the answer \bmod (10^9+7).

Examples

Input

4 2
1 1 3 4


Output

11


Input

6 3
10 10 10 -10 -10 -10


Output

360


Input

3 1
1 1 1


Output

0


Input

10 6
1000000000 1000000000 1000000000 1000000000 1000000000 0 0 0 0 0


Output

999998537
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e5+7;
const ll mod=1e9+7;
int T;
ll a[maxn];
int n,k;
ll f[maxn];
ll kqow(ll a,ll b)
{
    ll ans=1;
    while(b)
    {
        if(b%2) ans*=a;
        a*=a;
        ans%=mod;
        a%=mod;
        b/=2;
    }
    return ans;
}
ll ans;
int main()
{
    f[1]=1;
    f[0]=1;
    for(int i=2;i<=100002;i++)  f[i]=i*f[i-1]%mod;
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++)   scanf("%lld",&a[i]);
    sort(a+1,a+1+n);
    ll ab=f[k-1];
    for(int i=1;i<=n;i++)
    {
        if(i>=k)
        {
            ans+= kqow(ab * f[i-k]%mod ,mod-2)  *  f[i-1]  %mod*a[i];
            ans%=mod;
        }
        if(n-i>=k-1)
        {
            ans-=kqow(ab*f[n-i-k+1]%mod,mod-2)  *  f[n-i]   %mod*a[i];
            ans%=mod;
        }
    }


    printf("%lld\n",(ans+mod)%mod);
}

