Let's call beauty of an array b_1, b_2, …, b_n (n > 1) — min_{1 ≤ i < j ≤ n} |b_i - b_j|.

You're given an array a_1, a_2, … a_n and a number k. Calculate the sum of beauty over all subsequences of the array of length exactly k. As this number can be very large, output it modulo 998244353.

A sequence a is a subsequence of an array b if a can be obtained from b by deletion of several (possibly, zero or all) elements.

Input

The first line contains integers n, k (2 ≤ k ≤ n ≤ 1000).

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^5).

Output

Output one integer — the sum of beauty over all subsequences of the array of length exactly k. As this number can be very large, output it modulo 998244353.

Examples

Input


4 3
1 7 3 5


Output


8

Input


5 5
1 10 100 1000 10000


Output


9

Note

In the first example, there are 4 subsequences of length 3 — [1, 7, 3], [1, 3, 5], [7, 3, 5], [1, 7, 5], each of which has beauty 2, so answer is 8.

In the second example, there is only one subsequence of length 5 — the whole array, which has the beauty equal to |10-1| = 9.
import java.util.Arrays;
import java.util.Scanner;

//Certificate: https://codeforces.com/contest/1188/submission/105617476

public class ArrayBeauty     {
    public static int[][] f;
    
    public static void main(String[] args) 
    {
    	
    	int n, k;
    	int result = 0;
    	int mod = 998244353;
        int d, l;
        
        Scanner scan = new Scanner(System.in);
        
        n = scan.nextInt();
        k = scan.nextInt();
        int [] a = new int [n];
        int [][] f = new int [n+1][k+1];
        
        for (int i = 0; i < n; i++) {
            a[i] = scan.nextInt();
        }
        scan.close();
        
        Arrays.sort(a);
        for (d = 1; d * (k - 1) <= a[n-1]; d++) 
        {
            f[0][0]=1;
            l = 0;
            for(int i = 1; i <= n; i++)
            {
                while( a[l]+d <= a[i-1] )
                    l++;
                f[i][0]=f[i-1][0];
                for(int j = 1; (j <= i) && (j <= k); j++)
                    f[i][j]=(f[i-1][j]+f[l][j-1]) % mod;
            }
            result = (result + f[n][k]) % mod;
        }
        System.out.println(result);
    }

	
}
