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 mod = 998244353;
int f[1005][1005];
int n, k, ans;
int a[1005];
int search(int st, int en, int m, int i) {
  int p = (st + en) / 2;
  if (a[m] - a[st] < i) return 0;
  if (a[m] - a[p + 1] < i && a[m] - a[p] >= i) {
    return p;
  } else if (a[m] - a[p + 1] >= i) {
    return search(p + 1, en, m, i);
  } else if (a[m] - a[p] < i) {
    return search(st, p, m, i);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= (a[n] - a[1]) / (k - 1) + 1; i++) {
    f[0][0] = 1;
    for (int m = 1; m <= n; m++) {
      f[m][0] = 1;
      int p = search(1, m, m, i);
      for (int j = 1; j <= k && j <= m; j++) {
        f[m][j] = f[m - 1][j] + f[p][j - 1];
        f[m][j] %= mod;
      }
    }
    ans += f[n][k] % mod;
    ans %= mod;
  }
  cout << ans;
}
