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>
const int N = 1005, mo = 998244353;
int n, k, a[N], f[N][N], i, j, l, p, ans;
inline void add(int& a, const int& b) {
  a += b - mo;
  a += a >> 31 & mo;
}
int main() {
  scanf("%d%d", &n, &k);
  for (i = 1; i <= n; ++i) scanf("%d", a + i);
  std::sort(a + 1, a + n + 1);
  for (i = 1; i <= 100000 / (k - 1); ++i) {
    for (j = 0; j <= n; ++j) f[0][j] = 1;
    for (j = 1; j <= k; ++j)
      for (l = 1, p = 0; l <= n; ++l) {
        for (; p < l && a[l] - a[p + 1] >= i; ++p)
          ;
        f[j][l] = f[j - 1][p];
        add(f[j][l], f[j][l - 1]);
      }
    add(ans, f[k][n]);
  }
  printf("%d\n", ans);
  return 0;
}
