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 main() {
  int n, k, sol = 0;
  cin >> n >> k;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a.begin(), a.end());
  int maxB = (a[n - 1] - a[0]) / (k - 1);
  vector<int> ble(maxB + 2);
  vector<vector<int>> b(n + 1);
  for (int i = 1; i <= maxB; i++) {
    b[0] = vector<int>(k + 1);
    b[0][0] = 1;
    int p = 0;
    for (int j = 1; j <= n; j++) {
      while (a[p] <= a[j - 1] - i) p++;
      b[j] = b[j - 1];
      for (int l = 0; l < k; l++) {
        b[j][l + 1] = (b[j][l + 1] + b[p][l]) % mod;
        if (b[j][l + 1] == 0) break;
      }
    }
    ble[i] = b[n][k];
  }
  for (int i = 0; i <= maxB; i++)
    sol = (sol + i * (long long)(ble[i] - ble[i + 1]) + mod) % mod;
  cout << sol << endl;
  return 0;
}
