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;
using LL = long long;
const int MOD = 998244353;
int f[1002][1000];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> x(n);
  for (int i = 0; i < n; ++i) cin >> x[i];
  sort(x.begin(), x.end());
  int ans = 0;
  for (int t = 1; t <= (x[n - 1] - x[0]) / (k - 1); ++t) {
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < k; ++j) f[i][j] = 0;
    }
    f[0][0] = 1;
    int pre = -1;
    for (int i = 1; i < n; ++i) {
      while (x[pre + 1] <= x[i] - t) ++pre;
      f[i][0] = i + 1;
      if (pre == -1) continue;
      for (int j = 1; j < k; ++j) {
        f[i][j] = f[i - 1][j] + f[pre][j - 1];
        if (f[i][j] >= MOD) f[i][j] -= MOD;
      }
    }
    ans += f[n - 1][k - 1];
    if (ans >= MOD) ans -= MOD;
  }
  cout << ans << endl;
  return 0;
}
