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 CONST = 998244353;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k;
  cin >> n >> k;
  vector<int> A(n);
  for (int i = 0; i < n; i++) {
    cin >> A[i];
  }
  int ans = 0;
  sort(A.begin(), A.end());
  vector<vector<int> > dpcnt(n, vector<int>(k + 1));
  vector<vector<int> > prefcnt(n, vector<int>(k + 1));
  for (int x = 1; x * (k - 1) <= A[n - 1]; x++) {
    for (int i = 0; i < n; i++) {
      for (int j = 0; j <= k; j++) {
        prefcnt[i][j] = dpcnt[i][j] = 0;
      }
    }
    for (int i = 0; i < n; i++) {
      dpcnt[i][1] = 1;
      if (i != 0) {
        prefcnt[i][1] = prefcnt[i - 1][1];
      }
      prefcnt[i][1] += dpcnt[i][1];
    }
    int j = 0;
    for (int i = 0; i < n; i++) {
      while (j < i && A[i] - A[j] >= x) {
        j++;
      }
      j--;
      if (j != -1) {
        for (int d = 1; d < k; d++) {
          dpcnt[i][d + 1] += prefcnt[j][d];
          prefcnt[i][d + 1] = prefcnt[i - 1][d + 1] + dpcnt[i][d + 1];
          if (prefcnt[i][d + 1] >= CONST) {
            prefcnt[i][d + 1] -= CONST;
          }
          if (dpcnt[i][d + 1] >= CONST) {
            dpcnt[i][d + 1] -= CONST;
          }
        }
      } else {
        j++;
      }
    }
    for (int i = 0; i < n; i++) {
      ans += dpcnt[i][k];
      if (ans >= CONST) {
        ans -= CONST;
      }
    }
  }
  cout << ans;
  return 0;
}
