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 a[1005], dp[1005][1005], dp_sum[1005][1005];
int main() {
  ios::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  int limit = a[n] / (k - 1);
  int ans = 0;
  for (int diff = 1; diff <= limit; diff++) {
    for (int j = 1; j <= n; j++) dp_sum[j][1] = j, dp[j][1] = 1;
    for (int ct = 2; ct <= k; ct++) {
      for (int j = 1, l = 1; j <= n; j++) {
        while (l < j && a[j] - a[l] >= diff) l++;
        dp[j][ct] = dp_sum[l - 1][ct - 1];
        dp_sum[j][ct] = dp[j][ct] + dp_sum[j - 1][ct];
        if (dp_sum[j][ct] >= 998244353) dp_sum[j][ct] -= 998244353;
      }
    }
    ans += dp_sum[n][k];
    if (ans >= 998244353) ans -= 998244353;
  }
  cout << ans << endl;
}
