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 MOD = 998244353;
int main() {
  int n, k;
  cin >> n >> k;
  int ais[n];
  for (int i = 0; i < n; i++) {
    cin >> ais[i];
  }
  sort(ais, ais + n);
  int ans = 0;
  int dp[k][n];
  for (int length = 0; length < k; length++) {
    for (int el = 0; el < n; el++) {
      dp[length][el] = 0;
    }
  }
  for (int x = 1; x < *max_element(ais, ais + n) / (k - 1) + 1; x++) {
    for (int length = 2; length < k + 1; length++) {
      int p = 0;
      if (length == 2) {
        for (int el = 1; el < n; el++) {
          while (ais[p] + x <= ais[el]) {
            p++;
          }
          dp[length - 1][el] = (dp[length - 1][el - 1] + p) % MOD;
        }
      } else {
        for (int el = 1; el < n; el++) {
          while (ais[p + 1] + x <= ais[el]) {
            p++;
          }
          dp[length - 1][el] =
              (dp[length - 1][el - 1] + dp[length - 2][p]) % MOD;
        }
      }
    }
    ans = (ans + dp[k - 1][n - 1]) % MOD;
  }
  cout << ans << endl;
}
