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 mod = 998244353;
int n, m, a[1010], d[1010][1010], dp[1010][1010];
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  sort(a, a + n);
  int ans = 0;
  for (int j = 0; j < n; j++) dp[1][j] = j;
  int X = 100000 / (m - 1) + 1;
  for (int t = 1; t < X; t++) {
    for (int i = 2; i <= m; i++) {
      for (int j = 0, k = 0; j < n; j++) {
        while (a[j] - a[k] >= t) k++;
        dp[i][j + 1] = dp[i][j] + dp[i - 1][k];
        if (dp[i][j + 1] >= mod) dp[i][j + 1] -= mod;
      }
    }
    ans += dp[m][n];
    if (ans >= mod) ans -= mod;
  }
  printf("%d\n", ans);
}
