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 a, b, c, d, e, f[1009], dp[1009][1009], lf, sm, pas, k,
    mod = 998244353;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> a >> k;
  for (c = 1; c <= a; c++) cin >> f[c];
  sort(f + 1, f + a + 1);
  for (d = 1; d <= f[a]; d++) {
    if (d * (k - 1) > f[a]) break;
    for (c = 1; c <= a; c++) {
      for (e = 1; e <= k; e++) {
        dp[c][e] = 0;
      }
    }
    for (e = 1; e <= k; e++) {
      lf = e - 1;
      sm = 0;
      for (c = e; c <= a; c++) {
        if (e == 1) {
          dp[c][e] = 1;
          continue;
        }
        while (lf < c && f[c] - f[lf] >= d) {
          sm += dp[lf][e - 1];
          sm %= mod;
          lf++;
        }
        dp[c][e] = sm;
        if (e == k) {
          pas += dp[c][k];
          pas %= mod;
        }
      }
    }
  }
  cout << pas;
  return 0;
}
