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>
#define inf 1000000007
using namespace std;
typedef long long ll;
int n,k;
ll a[100009],b[100009],c[100009],ans=0;
ll pow1(ll x,ll y)
{
    ll res=1;
    while(y)
    {
        if(y%2)
            res=res*x%inf;
        x=x*x%inf;
        y/=2;
    }
    return res;
}
ll inv(ll x)
{
    return pow1(x,inf-2);
}
ll f(ll x,int q,int w,int e)
{
    return x*b[q]%inf*c[w]%inf*c[e]%inf;
}
int main()
{
    scanf("%d%d",&n,&k);
    b[0]=1,c[0]=1;
    for(int i=1; i<=n-1; i++)
        b[i]=b[i-1]*(ll)i%inf,c[i]=inv(b[i]);
    for(int i=1; i<=n; i++)
        scanf("%lld",&a[i]);
    sort(a+1,a+n+1);
    for(int i=k; i<=n; i++)
        ans=(ans+f(a[i],i-1,k-1,i-k))%inf;
    for(int j=1; j<=n-k+1; j++)
        ans=(ans-f(a[j],n-j,k-1,n-j-k+1)+inf)%inf;
    printf("%lld",ans);
    return 0;
}
