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 MAXN = 1005;
const int MOD = 998244353;
using namespace std;
int N, K, a[MAXN], cnt[100005];
int dp[MAXN][MAXN];
int do_dp_stuff(int dist) {
  for (int i = 0; i < N; i++) {
    dp[1][i] = 1;
  }
  for (int k = 2; k <= K; k++) {
    int ptr = N - 1, num = 0;
    for (int i = N - 1; i >= 0; i--) {
      while (ptr > i && a[ptr] - a[i] >= dist) {
        num = (num + dp[k - 1][ptr]) % MOD;
        ptr--;
      }
      dp[k][i] = num;
    }
  }
  int tot = 0;
  for (int i = 0; i < N; i++) {
    tot = (tot + dp[K][i]) % MOD;
  }
  return tot;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> K;
  for (int i = 0; i < N; i++) {
    cin >> a[i];
  }
  sort(a, a + N);
  int lim = (a[N - 1] - a[0]) / (K - 1), ans = 0;
  for (int i = 0; i <= lim + 1; i++) {
    cnt[i] = do_dp_stuff(i);
    if (i > 0) {
      ans = (ans + (long long)(i - 1) * (cnt[i - 1] - cnt[i])) % MOD;
    }
  }
  cout << ans << '\n';
  return 0;
}
