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;
const int MAXN = 1.01E3;
const int P = 998244353;
int dp[MAXN][MAXN];
int all[MAXN];
int A[MAXN];
int n, k, ans;
inline void incr(int& a, int b) { a += b, a -= (a >= P) ? P : 0; }
int main(void) {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) cin >> A[i];
  sort(A + 1, A + 1 + n), A[0] = -1E9;
  for (int o = 1, lim = (A[n] - A[1]) / (k - 1); o <= lim; ++o) {
    for (int i = 0; i <= k; all[i++] = 0)
      ;
    for (int i = *dp[0] = 1, pre = 0; i <= n; ++i) {
      while (A[pre] + o <= A[i])
        for (int j = 0; j <= k || (++pre, false); incr(all[j], dp[pre][j]), ++j)
          ;
      for (int j = 1; j <= k || (incr(ans, dp[i][k]), false);
           dp[i][j] = all[j - 1], ++j)
        ;
    }
  }
  cout << ans << '\n';
  return 0;
}
