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>
int N, A[1000], K;
static constexpr int MOD = 998244353;
int subtask(int D) {
  static int dp[1001][1001];
  for (int i = 0; i <= N; ++i) dp[i][1] = i;
  for (int i = 2; i <= K; ++i) {
    for (int j = 0; j < i; ++j) dp[j][i] = 0;
    for (int j = i; j <= N; ++j) {
      dp[j][i] = dp[j - 1][i];
      int *k = std::upper_bound(A, A + N, A[j - 1] - D);
      if (k == A) continue;
      --k;
      dp[j][i] += dp[(k - A) + 1][i - 1];
      dp[j][i] %= MOD;
    }
  }
  return dp[N][K];
}
int main() {
  scanf("%d%d", &N, &K);
  for (int i = 0; i < N; ++i) scanf("%d", &A[i]);
  std::sort(A, A + N);
  int ans = 0;
  for (int d = 1;; ++d) {
    int k = subtask(d);
    if (k == 0) break;
    ans += k;
    ans %= MOD;
  }
  printf("%d\n", ans);
  return 0;
}
