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<algorithm>
#define mod 1000000007
int n,m,f[100005];
long long fac[100005],inv[100005],ans;
long long c(int x,int y){return y<=x?fac[x]*inv[y]%mod*inv[x-y]%mod:0;}
long long fp(long long x,long long y){return (y%2?x:1ll)*(y/2?fp(x*x%mod,y/2):1ll)%mod;}
void pre(int n)
{
	fac[0]=inv[0]=1;
	for (int i=1;i<=n;i++) fac[i]=fac[i-1]*i%mod;
	for (int i=1;i<=n;i++) inv[i]=fp(fac[i],mod-2);
}
int main()
{
	scanf("%d%d",&n,&m);
	pre(std::max(n,m));
	for (int i=1;i<=n;i++) scanf("%d",&f[i]);
	std::sort(f+1,f+n+1);
	for (int i=1;i<=n;i++) ans=(ans+f[i]*c(i-1,m-1))%mod;
	for (int i=1;i<=n;i++) ans=(ans-f[i]*c(n-i,m-1)+mod)%mod;
	printf("%lld\n",ans);
	return 0;
}