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>
const int mod = 998244353;
const int inf = 1000000009;
const long long int INF = 1000000000000000009;
const long long big = 1000000000000000;
const long double eps = 0.0000000001;
using namespace std;
int maks = 0;
int T[1005];
int dp[1005][1005];
int main() {
  ios::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> T[i];
  sort(T + 1, T + n + 1);
  maks = T[n];
  long long int wynik = 0;
  for (int i = 1; i * (m - 1) <= maks; i++) {
    for (int j = 0; j <= m; j++)
      for (int k = 1; k <= n; k++) dp[j][k] = 0;
    for (int j = 1; j <= n; j++) dp[1][j] = 1;
    for (int j = 2; j <= m; j++) {
      int wsk = 1;
      long long int x = 0;
      for (int k = 1; k <= n; k++) {
        while (wsk < k && T[k] - T[wsk] >= i) {
          x += dp[j - 1][wsk];
          wsk++;
        }
        x %= mod;
        dp[j][k] = x;
      }
    }
    for (int j = 1; j <= n; j++) wynik += dp[m][j];
    wynik %= mod;
  }
  cout << wynik;
  return 0;
}
