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 mod = 998244353;
int n, k, a[1003], r, dp[1003][1003], dpp[1003], fp;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  sort(a + 1, a + n + 1);
  dpp[0] = 1;
  for (int i = 1; i <= (a[n] - a[1]) / (k - 1); ++i) {
    r = 0;
    for (int j = 1; j <= k; ++j) dpp[j] = 0;
    for (int j = 1; j <= n; ++j) {
      while (a[j] - a[r + 1] >= i) {
        ++r;
        for (int ii = 1; ii <= k; ++ii) {
          (dpp[ii] += dp[r][ii]) %= mod;
        }
      }
      for (int ii = 1; ii <= k; ++ii) {
        dp[j][ii] = dpp[ii - 1];
      }
      (fp += dp[j][k]) %= mod;
    }
  }
  cout << fp << endl;
  return 0;
}
