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], Max;
long long f[1005][1005][2], ans;
int main() {
  scanf("%d%d", &n, &K);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    Max = max(Max, a[i]);
  }
  sort(a + 1, a + n + 1);
  Max = Max / (K - 1);
  for (int i = 1; i <= Max + 5; i++) {
    f[1][1][0] = 1;
    int l = 1;
    for (int j = 2; j <= n; j++) {
      f[j][1][0] = f[j - 1][1][0] + 1;
      f[j][1][1] = 0;
      for (; a[j] - a[l] > i; l++)
        ;
      for (int k = 2; k <= min(K, j); k++) {
        f[j][k][0] = (f[j - 1][k][0] + f[l - 1][k - 1][0]) % 998244353;
        f[j][k][1] = (f[j - 1][k][1] + f[l - 1][k - 1][1]) % 998244353;
      }
      int L = l;
      for (; a[j] - a[l] == i; l++)
        ;
      if (l != L)
        for (int k = 2; k <= min(K, j); k++)
          f[j][k][1] = (f[j][k][1] + f[l - 1][k - 1][0] - f[L - 1][k - 1][0] +
                        f[l - 1][k - 1][1] - f[L - 1][k - 1][1]) %
                       998244353;
      l = L;
    }
    ans = (ans + f[n][K][1] * i % 998244353) % 998244353;
  }
  ans = (ans + 998244353) % 998244353;
  printf("%lld\n", ans);
}
