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>
using namespace std;
const int MAXN=100010,P=1e9+7;
int n,k,ans,a[MAXN],b[MAXN],c[MAXN],fac[MAXN],inv[MAXN];
int qpow (int a,int b) {
	int res=1;
	while (b) {
		if (b&1) {res=(1ll*res*a)%P;}
		a=(1ll*a*a)%P,b>>=1;
	}
	return res;
}
void init () {
	fac[0]=inv[0]=1;
	for (int i=1;i<=MAXN-10;i++) {fac[i]=(1ll*fac[i-1]*i)%P;}
	inv[MAXN-10]=qpow(fac[MAXN-10],P-2);
	for (int i=MAXN-11;i>=1;i--) {inv[i]=(1ll*inv[i+1]*(i+1))%P;}
	return;
}
int ct (int n,int m) {return (n<m?0:(1ll*((1ll*fac[n]*inv[m])%P)*inv[n-m])%P);}
int main () {
	scanf("%d%d",&n,&k);
	for (int i=1;i<=n;i++) {scanf("%d",&a[i]);}
	sort(a+1,a+n+1);
	a[0]=a[n+1]=1e9+1;
	for (int i=1;i<=n;i++) {b[i]=i-1;c[i]=n-b[i]-1;}
	init();
	for (int i=1;i<=n;i++) {ans=((ans+(1ll*a[i]*ct(b[i],k-1))%P)%P+P)%P;}
	for (int i=1;i<=n;i++) {ans=((ans-(1ll*a[i]*ct(c[i],k-1))%P)%P+P)%P;}
	printf("%d\n",ans);
	return 0;
}