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 <iostream>
using namespace std;
#include <algorithm>
#define ll long long
ll a[100005],sum[100005],jc[100005];
const ll mod=1e9+7;
ll app(ll a,ll b)
{
	return ((a%mod)*(b%mod))%mod;
}
ll ksm(ll a,ll b)
{
	ll ret=1;
	while(b)
	{
		if(b%2==1)ret=app(ret,a);
		a=app(a,a);
		b/=2;
	}
	return ret;
}
ll C(ll a,ll b)
{
	if(b==0)return 1;
	return app(jc[a],app(ksm(jc[b],mod-2),ksm(jc[a-b],mod-2)));
}
int main()
{
	int n,k;
	scanf("%d%d",&n,&k);
	jc[0]=1;
	for(int i=1;i<=n;i++)
	jc[i]=app(jc[i-1],i);

	for(int i=1;i<=n;i++)scanf("%lld",&a[i]);
	sort(a+1,a+n+1);
	for(int i=1;i<=n;i++)
	sum[i]=sum[i-1]+a[i];
	ll ans=0;
	for(int t=k;t<=n;t++)
	{
		ll len=t-1;
		ll now=(-sum[n-len]+sum[n]-sum[len])%mod;
		//cout<<now<<" "<<C(t-2,k-2)<<endl;
		ans=(ans%mod+app(C(t-2,k-2),now)%mod)%mod;
	
	}
	cout<<ans;
}