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;
const int MAXA = 2e5;
int n, K, a[MAXA], last, f[2000][2000], ans;
const int P = 998244353;
int main() {
  scanf("%d%d", &n, &K);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= (a[n] - a[1]) / (K - 1); i++) {
    last = 1;
    for (int j = 1; j <= n; j++) {
      for (int k = 1; k <= K; k++) {
        while (a[j] - a[last] >= i) last++;
        if (last) last--;
        f[j][k] = (f[last][k - 1] + f[j - 1][k]) % P;
      }
      f[j][1]++;
    }
    ans = (ans + f[n][K]) % P;
  }
  printf("%d\n", ans);
}
