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>
int compare(const void *a, const void *b) {
  int ia = *(int *)a;
  int ib = *(int *)b;
  return ia - ib;
}
int main() {
  static int aa[1000], dp[1000];
  int n, k, x, h, i, j, ans;
  scanf("%d%d", &n, &k);
  for (i = 0; i < n; i++) scanf("%d", &aa[i]);
  qsort(aa, n, sizeof *aa, compare);
  ans = 0;
  for (x = 1; x * (k - 1) <= aa[n - 1] - aa[0]; x++) {
    for (i = 0; i < n; i++) dp[i] = 1;
    for (h = 1; h < k; h++) {
      for (i = 1; i < n; i++) dp[i] = (dp[i] + dp[i - 1]) % 998244353;
      for (i = j = n - 1; j >= 0; j--) {
        while (i >= 0 && aa[j] - aa[i] < x) i--;
        dp[j] = i >= 0 ? dp[i] : 0;
      }
    }
    for (i = 0; i < n; i++) ans = (ans + dp[i]) % 998244353;
  }
  printf("%d\n", ans);
  return 0;
}
