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 mo(int x) {
  if (x < 0) x += mod;
  if (x >= mod) x -= mod;
  return x;
}
int dp[1005][1005], sum[1005][1005], summ[1005][1005], a[1005], cf[1005], n, K;
signed main() {
  cin >> n >> K;
  if (K == 1) {
    puts("0");
    return 0;
  }
  for (int i = 1; i <= n; i++) cin >> a[i];
  a[0] = -1e9;
  sort(a + 1, a + n + 1);
  int ans = 0, maxn = a[n] / (K - 1);
  for (int i = 1; i <= maxn; i++) {
    dp[0][0] = sum[0][0] = 1;
    int tmp = 0;
    for (int j = 1; j <= n; j++) {
      while (a[j] - a[tmp] >= i) tmp++;
      sum[j][0] = 1;
      for (int k = 1; k <= K; k++) {
        dp[j][k] = sum[tmp - 1][k - 1];
        sum[j][k] = mo(sum[j - 1][k] + dp[j][k]);
      }
      ans = mo(ans + dp[j][K]);
    }
  }
  cout << ans;
}
