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 ll long long
using namespace std;
ll n,k;
ll ans,mod=1000000007,c[100010],a[100010],f[100010];
ll po(ll x,ll p)
{
    if(p==0)
    {
        return 1;
    }
    ll ret=1;
    if(p%2)
    {
        ret=x;
        ret%=mod;
    }
    ll tem=po(x,p/2);
    ret*=tem;
    ret%=mod;
    ret*=tem;
    ret%=mod;
    return ret;
}
int main()
{
    scanf("%lld%lld",&n,&k);
    f[0]=1;
    for(int i=1;i<=n;i++)
    {
        scanf("%lld",a+i);
        f[i]=f[i-1]*i;
        f[i]%=mod;
    }
    sort(a+1,a+n+1);
    for(ll i=k-1;i<=n;i++)
    {
        c[i]=f[i]*po(f[k-1],mod-2);
        c[i]%=mod;
        c[i]*=po(f[i-k+1],mod-2);
        c[i]%=mod;
    }
    for(int i=1;i<=n;i++)
    {
        ans-=c[n-i]*a[i];
        ans+=mod*1000000000;
        ans%=mod;
        ans+=c[i-1]*a[i];
        ans%=mod;
    }
    printf("%lld",ans);
}