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[1000];
int dp[1000][1001];
int calc(int x) {
  int p = 0;
  dp[0][0] = dp[0][1] = 1;
  for (int i = 1; i < n; i++) {
    while (a[i] - a[p + 1] >= x) p++;
    dp[i][0] = 1;
    dp[i][1] = i + 1;
    for (int j = 2; j <= k; j++) {
      dp[i][j] = dp[i - 1][j];
      if (a[i] - a[p] >= x) dp[i][j] = (dp[i][j] + dp[p][j - 1]) % 998244353;
    }
  }
  return dp[n - 1][k];
}
int main(void) {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  sort(a, a + n);
  int maxi = a[n - 1] / (k - 1), ans = 0;
  for (int i = 1; i <= maxi; i++) ans = (ans + calc(i)) % 998244353;
  printf("%d\n", ans);
  return 0;
}
