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.
#include <bits/stdc++.h>
using namespace std;
int N, K, A[1005];
int f[1005][1005], s[1005][1005];
int main() {
  scanf("%d%d", &N, &K);
  for (int i = 1; i <= N; i++) scanf("%d", &A[i]);
  sort(A + 1, A + N + 1);
  int ans = 0;
  for (int L = 1; L <= 100000 / (K - 1); L++) {
    int j = 0;
    f[0][0] = s[0][0] = 1;
    for (int i = 1; i <= N; i++) {
      while (j + 1 < i && A[i] - A[j + 1] >= L) j++;
      s[i][0] = 1;
      for (int k = 1; k <= K; k++)
        f[i][k] = s[j][k - 1], s[i][k] = (s[i - 1][k] + f[i][k]) % 998244353;
      ans = (ans + f[i][K]) % 998244353;
    }
  }
  printf("%d\n", ans);
}
