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 = 1e9 + 7;
int as[1005] = {-INF};
int N, K;
int dp[1005][1005] = {{1}};
int solve(int crit) {
  int j = 0;
  for (int i = 1; i <= N; i++) {
    for (int k = 0; k <= K; k++) {
      dp[i][k] = dp[i - 1][k];
    }
    while (as[j + 1] <= as[i] - crit) j++;
    for (int k = 0; k < K; k++) {
      dp[i][k + 1] = (dp[i][k + 1] + dp[j][k]) % MOD;
    }
  }
  return dp[N][K];
}
int main() {
  scanf("%d %d", &N, &K);
  for (int i = 1; i <= N; i++) {
    scanf("%d", &as[i]);
  }
  std::sort(as + 1, as + N + 1);
  int ans = 0;
  for (int i = 1; i * (K - 1) <= 100000; i++) {
    ans = (ans + solve(i)) % MOD;
  }
  printf("%d\n", ans);
  return 0;
}
