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 N = 1e3 + 10, mod = 998244353;
int n, k, ans, a[N], dp[N][N];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  for (int nd = 1; nd * (k - 1) <= a[n - 1]; nd++) {
    for (int i = 0; i <= n; i++)
      for (int j = 0; j <= k; j++) dp[i][j] = 0;
    int plc = -1;
    dp[0][0] = 1;
    for (int i = 1; i <= n; i++) {
      while (a[i - 1] - a[plc + 1] >= nd) plc++;
      for (int j = 0; j <= k; j++) {
        if (j == 0) {
          dp[i][j] = dp[i - 1][j];
        } else {
          dp[i][j] = (dp[i - 1][j] + dp[plc + 1][j - 1]) % mod;
        }
      }
    }
    ans = (ans + dp[n][k]) % mod;
  }
  cout << ans;
  return 0;
}
