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;
#define rep(i,n) for(int i=0; i<(n); i++)
using ll = long long;
constexpr int MX = 100005, MOD = 1e9+7;
int n, k, a[MX];
ll fact[MX] = {1}, invf[MX], ans;
ll exp(ll b, ll n){
	ll res = 1;
	for(; n; n/=2,(b*=b)%=MOD) if(n%2) (res *= b) %= MOD;
	return res;
}
void setcomb(int n){
	for(int i = 1; i <= n; i++) fact[i] = (fact[i-1] * i) % MOD;
	invf[n] =  exp(fact[n] ,MOD-2);
	for(int i = n; i > 0; i--) invf[i-1] = (invf[i] * i) % MOD;
}
inline ll comb(int a, int b){return fact[a] * invf[a-b] % MOD * invf[b] % MOD;}
int main(){
	scanf("%d%d", &n, &k);
	rep(i,n) scanf("%d", a+i);
	sort(a,a+n);
	setcomb(n);
	rep(i,n-k+1) ans -= a[i] * comb(n-i-1,k-1) % MOD;
	for(int i=n-1; i>=k-1; i--) ans += a[i] * comb(i,k-1) % MOD;
	printf("%lld\n", (ans % MOD + MOD) % MOD);
}