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;
vector<pair<int, int> > sol;
long long ar[1005], DP[1005][1005], pre[1005][1005];
long long N, K;
void solve(int x) {
  for (int i = 0; i < N; i++)
    for (int j = 0; j <= K; j++) {
      DP[i][j] = 0;
      pre[i][j] = 0;
    }
  for (int i = 0; i < N; i++) DP[i][1] = 1;
  pre[0][1] = 1;
  int lft = 0;
  for (int i = 0; i < N; i++) {
    while (ar[i] - ar[lft] >= x && lft <= (i - 1)) lft++;
    lft--;
    if (ar[i] - ar[0] < x) {
      for (int j = 1; j <= K; j++) {
        pre[i][j] = (i == 0 ? 0 : pre[i - 1][j]) + DP[i][j];
        pre[i][j] %= 998244353;
      }
      continue;
    }
    for (int j = 1; j <= K; j++) {
      DP[i][j] += (pre[lft][j - 1]);
      DP[i][j] %= 998244353;
    }
    for (int j = 1; j <= K; j++) {
      pre[i][j] = (i == 0 ? 0 : pre[i - 1][j]) + DP[i][j];
      pre[i][j] %= 998244353;
    }
  }
}
int main() {
  cin >> N >> K;
  for (int i = 0; i < N; i++) cin >> ar[i];
  sort(ar, ar + N);
  long long ans = 0ll;
  for (int i = 1; i <= ((100000 / (K - 1)) + 5); i++) {
    solve(i);
    ans += pre[N - 1][K];
    ans %= 998244353;
  }
  cout << ans << endl;
  return (0);
}
