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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int maxn=100010,mod=1000000007;
typedef long long ll;
ll s[maxn],ans;
int n,k,a[maxn],inv[maxn],fac[maxn];

int c(int x,int y)
{
	return 1ll*fac[x]*inv[y]%mod*inv[x-y]%mod;
}

int main()
{
	int i,j;
	scanf("%d%d",&n,&k);
	if (k==1||n<k)
	{
		puts("0");
		return 0;
	}
	for (i=1;i<=n;i++)
		scanf("%d",&a[i]);
	sort(a+1,a+n+1);
	if (n==k)
	{
		printf("%d\n",(a[n]-a[1])%mod);
		return 0;
	}
	for (i=1;i<=n;i++)
		s[i]=s[i-1]+1ll*a[i];
	inv[0]=inv[1]=fac[0]=fac[1]=1;
	for (i=2;i<=n;i++)
		inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
	for (i=2;i<=n;i++)
	{
		inv[i]=1ll*inv[i-1]*inv[i]%mod;
		fac[i]=1ll*fac[i-1]*i%mod;
	}
	for (i=k-2;i<=n-2;i++)
	{
		ans=(ans+1ll*c(i,k-2)*(((s[n]-s[i+1]-s[n-i-1])%mod+mod)%mod)%mod)%mod;
	}
	printf("%d\n",ans);
	return 0;
}
