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;
long long ans;
long long n;
long long k;
long long a[1005];
long long f[1005][1005];
long long s[1005][1005];
int main() {
  scanf("%lld %lld", &n, &k);
  for (long long i = 1; i <= n; ++i) scanf("%lld", a + i);
  sort(a + 1, a + 1 + n);
  s[0][0] = 1LL;
  for (long long gap = 1; gap * (k - 1) <= a[n] - a[1]; ++gap) {
    long long pre = 0;
    for (long long i = 1; i <= n; ++i) {
      while (a[i] - a[pre + 1] >= gap) ++pre;
      f[i][1] = 1LL;
      s[i][1] = (s[i - 1][1] + f[i][1]) % 998244353;
      for (long long j = 2; j <= min(i, k); ++j) {
        f[i][j] = s[pre][j - 1];
        s[i][j] = (s[i - 1][j] + f[i][j]) % 998244353;
      }
    }
    ans = (ans + s[n][k]) % 998244353;
  }
  printf("%lld\n", ans);
  return 0;
}
