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;
int a[1010];
int dp[1010][1010];
int main() {
  while (scanf("%d%d", &n, &k) != EOF) {
    memset(dp, 0, sizeof(dp));
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
    }
    sort(a + 1, a + 1 + n);
    long long ans = 0;
    for (int i = a[n] / (k - 1); i >= 1; i--) {
      int l = 1;
      for (int j = 1; j <= n; j++) {
        while (l < j && a[j] >= a[l] + i) {
          l++;
        }
        dp[j][2] = (dp[j - 1][2] + l - 1) % 998244353;
        for (int m = 3; m <= k; m++) {
          dp[j][m] = (dp[l - 1][m - 1] + dp[j - 1][m]) % 998244353;
        }
      }
      ans = (ans + dp[n][k]) % 998244353;
    }
    printf("%lld\n", ans);
  }
  return 0;
}
