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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, k;
  cin >> n >> k;
  vector<int> a((size_t)n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  int m = *max_element(a.begin(), a.end());
  sort(a.begin(), a.end());
  int o = 0;
  for (int x = 1; x <= m / (k - 1) + 1; x++) {
    vector<vector<int>> dp((size_t)k + 1, vector<int>((size_t)n, 0));
    vector<vector<int>> ep((size_t)k + 1, vector<int>((size_t)n, 0));
    int r = 0;
    for (int i = 0; i < n; i++) {
      while (r < i && a[i] >= a[r] + x) {
        r++;
      }
      for (int j = 1; j <= k; j++) {
        if (j == 1) {
          dp[j][i] = 1;
        } else {
          dp[j][i] += r == 0 ? 0 : ep[j - 1][r - 1];
        }
        dp[j][i] %= 998244353;
        ep[j][i] = (i == 0 ? 0 : ep[j][i - 1]) + dp[j][i];
        ep[j][i] %= 998244353;
      }
    }
    o += ep[k][n - 1];
    o %= 998244353;
  }
  cout << o << endl;
  return 0;
}
