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 n, k;
long long cntsubs[1001][1001];
long long prefsubs[1001][1001];
long long a[1001];
long long dp(int index, int elemleft, int dist) {
  if (elemleft == 0)
    return 1;
  else if (index == 0)
    return 0;
  if (cntsubs[index][elemleft] == -1) {
    cntsubs[index][elemleft] = 0;
    for (int i = 0; i < index; i++) {
      if (a[i] + dist <= a[index])
        cntsubs[index][elemleft] += dp(i, elemleft - 1, dist),
            cntsubs[index][elemleft] %= 998244353;
    }
  }
  return cntsubs[index][elemleft];
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  long long ans = 0;
  for (int i = 1; i <= 100000 / (k - 1); i++) {
    for (int j = 0; j < n; j++) {
      cntsubs[j][0] = 1;
      prefsubs[j][0] = j + 1;
    }
    for (int j = 1; j < k; j++) {
      long long bestind = 0;
      for (int l = 0; l < n; l++) {
        while (a[bestind] + i <= a[l]) bestind++;
        cntsubs[l][j] = (bestind != 0 ? prefsubs[bestind - 1][j - 1] : 0);
      }
      prefsubs[0][j] = cntsubs[0][j];
      for (int l = 1; l < n; l++) {
        prefsubs[l][j] = cntsubs[l][j] + prefsubs[l - 1][j];
        prefsubs[l][j] %= 998244353;
      }
    }
    ans += prefsubs[n - 1][k - 1];
    ans %= 998244353;
  }
  cout << ans << '\n';
}
