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, a[1010];
long long f[1010][1010];
const long long mo = 998244353LL;
void add(long long &x, long long y) {
  x += y;
  if (x > mo) x %= mo;
}
int main() {
  scanf("%d%d", &n, &k);
  int Max = 0;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), Max = max(Max, a[i]);
  sort(a + 1, a + 1 + n);
  Max /= k - 1;
  long long ans = 0;
  for (int bea = 1; bea <= Max; bea++) {
    for (int i = 1; i <= k; i++)
      for (int j = 1; j <= n; j++) f[i][j] = 0;
    for (int i = 1; i <= n; i++) f[1][i] = 1;
    for (int i = 2; i <= k; i++) {
      long long sum = 0;
      int l = 1;
      for (int j = 1; j <= n; j++) {
        while (l < j && a[j] - a[l] >= bea) add(sum, f[i - 1][l]), l++;
        add(f[i][j], sum);
      }
    }
    for (int i = 1; i <= n; i++) add(ans, f[k][i]);
  }
  cout << ans << endl;
  return 0;
}
