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
import java.util.*;
import java.lang.*;

class Mod{
    static final long MOD = 1000_000_007;
    static final long[] fact = makeFactorialArray(1000000);
    static final long[] factInv = makeInversedArray(fact);

    public static long add(long a, long b){
        long tmp = (a+b)%MOD;
        return tmp>=0 ? tmp : tmp+MOD;
    }
    public static long adds(long... ar){
        long ans=0;
        for(long x:ar) ans = Mod.add(ans, x);
        return ans;
    }
    public static long sub(long a, long b){
        return add(a,-b);
    }
    public static long mult(long a, long b){
        long tmp = (a*b)%MOD;
        return tmp>=0 ? tmp : tmp+MOD;
    }
    public static long mults(long... ar){
        long ans=1;
        for(long x:ar) ans = Mod.mult(ans, x);
        return ans;
    }
    public static long power(long a, long x){ //calculate a^x
        if(x<0)return 0;
        if(x==0)return 1;
        if(x%2==0) {
            long half = power(a,x/2);
            return mult(half,half);
        }
        return (a*power(a,x-1)) % MOD;
    }
    public static long inverse(long a){
        return power(a,MOD-2);
    }
    public static long div(long a, long b){
        return mult(a, inverse(b));
    }
    public static long[] makeFactorialArray(int size){
        long[] array = new long[size];
        array[0]=1;
        for(int i=1;i<size;i++){
            array[i]=mult(array[i-1],i);
        }
        return array;
    }
    public static long[] makeInversedArray(long[] original){
        long[] array = new long[original.length];
        for(int i=0;i<original.length;i++){
            array[i] = inverse(original[i]);
        }
        return array;
    }
    public static long combination(long n, long r){
        int N = (int)n,  R = (int)r;
        if(n<0 || r<0 || n<r) return 0;
        return mults(fact[N],factInv[R],factInv[N-R]);
    }
    public static long permutation(long n, long r){
        int N = (int)n,  R = (int)r;
        if(n<0 || r<0 || n<r) return 0;
        return mult(fact[N], factInv[N-R]);
    }
}
public class Main {

    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int K = sc.nextInt();
        long[] A = new long[N];
        for(int n=0; n<N; n++) A[n]=sc.nextLong();
        Arrays.sort(A);

        long ans = 0;

        for(int n=0; n<N; n++){
            ans = Mod.add(ans,
                          Mod.mults(A[n],
                                    Mod.sub(Mod.combination(n, K-1),
                                            Mod.combination(N-n-1, K-1))));
        }

        System.out.println(ans);

    }

}
