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;
int mod = 998244353;
const int maxn = 1005;
int dp[maxn][maxn];
int data[maxn];
int sum(int p1, int p2) { return (p1 + p2) % mod; }
int mult(int p1, int p2) { return 1LL * p1 * p2 % mod; }
int sub(int p1, int p2) { return (((long long int)p1 - p2) % mod + mod) % mod; }
long long int n, k;
void solve() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> data[i];
  sort(data + 1, data + n + 1);
  int maxd = (data[n] - data[1]) / (k - 1);
  int ans = 0;
  for (int d = 1; d <= maxd; d++) {
    for (int i = 1; i <= n; i++) {
      int idx =
          lower_bound(data + 1, data + n, max(0, data[i] - d + 1)) - data - 1;
      dp[i][1] = i;
      for (int j = 2; j <= k; j++) {
        dp[i][j] = sum(dp[i - 1][j], dp[idx][j - 1]);
      }
    }
    ans = sum(ans, dp[n][k]);
  }
  cout << ans << endl;
}
int main() { solve(); }
