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 N = 1e3 + 2;
const int inf = 1e9 + 7;
const int mod = 998244353;
int ar[N], dp[N][N], sum[N][N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, i, j, k, l, num, max1 = 0, ans = 0;
  cin >> n >> num;
  for (i = 1; i <= n; i++) {
    cin >> ar[i];
    max1 = max(max1, ar[i]);
  }
  ar[0] = -inf;
  for (i = 0; i <= n; i++) {
    sum[0][i] = 1;
  }
  sort(ar + 1, ar + 1 + n);
  for (i = 1; i <= max1 / (num - 1) + 1; i++) {
    for (j = 1; j <= num; j++) {
      l = 0;
      for (k = 1; k <= n; k++) {
        while (l <= n && ar[l] + i <= ar[k]) {
          l++;
        }
        dp[j][k] = sum[j - 1][l - 1];
        sum[j][k] = sum[j][k - 1] + dp[j][k];
        if (sum[j][k] >= mod) {
          sum[j][k] -= mod;
        }
      }
    }
    ans += sum[num][n];
    if (ans >= mod) {
      ans -= mod;
    }
  }
  cout << ans;
}
