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 int64_t MOD = 998244353;
void add(int64_t& a, int64_t b) { a = (a + b) % MOD; }
void mul(int64_t& a, int64_t b) { a = a * b % MOD; }
int main() {
  int N, K, A[1000];
  cin >> N >> K;
  for (int i = 0; i < N; i++) cin >> A[i];
  sort(A, A + N);
  int64_t ans = 0;
  vector<vector<int64_t>> dp(N + 1, vector<int64_t>(K + 1));
  for (int x = 1; x * (K - 1) <= 100000; x++) {
    for (int i = 0; i <= N; i++) fill(dp[i].begin(), dp[i].end(), 0);
    dp[0][0] = 1;
    int L = 0;
    for (int i = 0; i < N; i++) {
      while (A[i] - A[L] >= x) L++;
      for (int k = 0; k <= K; k++) {
        add(dp[i + 1][k], dp[i][k]);
        if (k < K) add(dp[i + 1][k + 1], dp[L][k]);
      }
    }
    add(ans, dp[N][K]);
  }
  cout << ans << endl;
  return 0;
}
