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;
typedef long long ll;
typedef pair<int,int> pii;

const int maxn = 1e5+23;
int a[maxn];
const int mod = 1e9+7;
int n,k;
int fac[maxn],inv[maxn];

int qpow(int a,int b)
{
    int ans=1;
    while(b){
        if(b&1) ans=1ll*ans*a%mod;
        b>>=1;
        a=1ll*a*a%mod;
    }
    return ans;
}

int comb(int i,int j)
{
    if(i<j) return 0;
    return 1ll*fac[i]*inv[j]%mod*inv[i-j]%mod;
}


int main()
{
    cin>>n>>k;
    fac[0]=1;
    for(int i=1;i<=n;i++) fac[i]=1ll*fac[i-1]*i%mod;
    inv[n]=qpow(fac[n],mod-2);
    for(int i=n-1;i>=0;i--) inv[i]=1ll*inv[i+1]*(i+1)%mod;
    for(int i=0;i<n;i++) scanf("%d",a+i);
    sort(a,a+n);ll ans=0;
    for(int i=n-1;i>=0;i--){
        ans=(ans+1ll*a[i]*comb(i,k-1)%mod)%mod;
    }
    for(int i=0;i<n;i++){
        ans=(ans-1ll*a[i]*comb(n-i-1,k-1)%mod+mod)%mod;
    }
    cout<<ans;
}
