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;
using LL = long long;
const int MX = 1005, mod = 998244353;
int n, k, a[MX], dp[MX][MX], ans;
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  a[0] = -100000;
  for (int i = 1; i <= n; i++) dp[i][1] = i;
  for (int D = 1; D * (k - 1) <= a[n] - a[1]; D++) {
    for (int l = 2; l <= k; l++) {
      for (int i = 1, j = 0; i <= n; i++) {
        while (a[j + 1] + D <= a[i]) ++j;
        dp[i][l] = (dp[i - 1][l] + dp[j][l - 1]) % mod;
      }
    }
    ans = (ans + dp[n][k]) % mod;
  }
  cout << ans << "\n";
  return 0;
}
