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], f[1005][1005], ans;
void add(int &x, int y) {
  x += y;
  if (x >= 998244353) x -= 998244353;
}
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 _ = (a[n] - a[1]) / (k - 1);
  for (int i = 0; i <= n; ++i) f[i][0] = 1;
  for (int i = 1; i <= _; ++i) {
    int l = 0;
    for (int r = 1; r <= n; ++r) {
      while (a[r] - a[l + 1] >= i) ++l;
      for (int j = 1; j <= k; ++j) {
        f[r][j] = f[l][j - 1] + f[r - 1][j];
        if (f[r][j] >= 998244353) f[r][j] -= 998244353;
      }
    }
    add(ans, f[n][k]);
  }
  printf("%d\n", ans);
  return 0;
}
