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 limit = 1005;
const long long int modulo = 998244353;
void add(long long int &a, long long int b) { a = (a + b) % modulo; }
int n, k;
int a[limit];
long long int memcompute[limit][limit];
long long int compute(int x) {
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= k; j++) memcompute[i][j] = 0;
  memcompute[0][0] = 1;
  int I = 0;
  for (int i = 1; i <= n; i++) {
    while (a[I + 1] + x <= a[i]) I++;
    for (int j = 0; j <= k; j++) {
      add(memcompute[i][j], memcompute[i - 1][j]);
      if (0 < j) add(memcompute[i][j], memcompute[I][j - 1]);
    }
  }
  return memcompute[n][k];
}
long long int sol;
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + (n + 1));
  for (long long int x = 1; x * (k - 1) <= a[n]; x++) add(sol, compute(x));
  cout << sol << endl;
}
