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 maxn = 1e6 + 10;
const long long mod = 998244353;
int n, m, a[1010], dp[2][1010], s[2][1010], ss[100010], h[1010], ans;
int main() {
  int i, j, k, ii, tp, ct;
  cin >> n >> m;
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1);
  for (k = 0; k <= a[n] - a[1]; k++) {
    for (ct = 1, tp = 1, i = 2; i <= n; i++) {
      if (a[i] - a[tp] >= k) {
        h[++ct] = i;
        tp = i;
      }
    }
    if (ct < m) break;
    for (i = 1; i <= n; i++) {
      dp[1][i] = 1;
      s[1][i] = i;
    }
    for (j = 2; j <= m; j++) {
      for (s[j % 2][h[j] - 1] = 0, ii = 0, i = h[j]; i <= n; i++) {
        while (ii + 1 <= i && a[i] - a[ii + 1] >= k) ii++;
        if (ii >= h[j - 1] && a[i] - a[ii] >= k)
          dp[j % 2][i] = s[1 - j % 2][ii];
        else
          dp[j % 2][i] = 0;
        s[j % 2][i] = (s[j % 2][i - 1] + dp[j % 2][i]) % mod;
      }
    }
    for (i = h[m]; i <= n; i++) ss[k] = (ss[k] + dp[m % 2][i]) % mod;
  }
  for (i = k - 1; i >= 0; i--)
    ans = (ans + (ss[i] + mod - ss[i + 1]) * i) % mod;
  cout << ans << "\n";
  return 0;
}
