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);
  int n, k;
  cin >> n >> k;
  int a[n + 1], temp, temp2;
  a[0] = -1000000;
  for (temp = 1; temp <= n; temp++) cin >> a[temp];
  sort(a + 1, a + n + 1);
  int ans = 0;
  for (int selisih = 1; selisih <= a[n] / (k - 1); selisih++) {
    int dp[k + 1][n + 1];
    memset(dp, 0, sizeof(dp));
    int pref[n + 1];
    pref[0] = 0;
    for (temp = 1; temp <= n; temp++) {
      dp[1][temp] = 1;
      pref[temp] = temp;
    }
    for (int length = 2; length <= k; length++) {
      int last_ptr = 0;
      for (int index = length; index <= n; index++) {
        while (a[index] - a[last_ptr + 1] >= selisih) last_ptr++;
        dp[length][index] = pref[last_ptr];
      }
      for (int index = 1; index <= n; index++) {
        pref[index] = pref[index - 1] + dp[length][index];
        if (pref[index] >= 998244353) pref[index] -= 998244353;
      }
    }
    for (int temp = 1; temp <= n; temp++) {
      ans += dp[k][temp];
      if (ans >= 998244353) ans -= 998244353;
    }
  }
  cout << ans << "\n";
}
