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;
int n, k, x[1010], dp[1010][1010], sm[1010];
int solve(int gap) {
  dp[0][0] = 1, x[0] = -200000;
  for (int i = 1; i <= k; ++i) {
    for (int j = 0; j <= n; ++j)
      sm[j] = j == 0 ? dp[i - 1][j] : (dp[i - 1][j] + sm[j - 1]) % 998244353;
    int pos = 0;
    for (int j = 1; j <= n; ++j) {
      while (x[j] - x[pos + 1] >= gap) pos++;
      dp[i][j] = sm[pos];
    }
  }
  int sum = 0;
  for (int i = 1; i <= n; ++i) sum = (sum + dp[k][i]) % 998244353;
  return sum;
}
int32_t main() {
  cin.tie(0), cout.sync_with_stdio(0);
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) cin >> x[i];
  sort(x + 1, x + 1 + n);
  int dis = 100000 / (k - 1), pre = solve(1), ans = 0;
  for (int i = 2; i <= dis + 5; ++i) {
    int now = solve(i);
    ans = (ans + 1ll * (pre - now + 998244353) * (i - 1)) % 998244353;
    pre = now;
  }
  cout << ans << "\n";
  return 0;
}
