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;
long long mod = 998244353;
int seq[1010];
long long dp[1010][1010];
void add(long long &x, long long y) { x = (x + y) % mod; }
int main() {
  int n, k, mmax = 0;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> seq[i];
    mmax = max(mmax, seq[i]);
  }
  sort(seq + 1, seq + 1 + n);
  mmax /= k - 1;
  long long ans = 0;
  for (int x = 1; x <= mmax; x++) {
    for (int i = 0; i <= k; i++)
      for (int j = 0; j <= n; j++) dp[i][j] = 0;
    for (int i = 1; i <= n; i++) dp[1][i] = 1;
    for (int i = 2; i <= k; i++) {
      long long sum = 0;
      int p = 1;
      for (int j = 1; j <= n; j++) {
        while (j > p && seq[j] - seq[p] >= x) {
          add(sum, dp[i - 1][p]);
          p++;
        }
        add(dp[i][j], sum);
      }
    }
    for (int i = 1; i <= n; i++) add(ans, dp[k][i]);
  }
  cout << ans << endl;
}
