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 dp[1005][2];
int arr[1005];
void solve() {
  int n, k;
  int mx = 0;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
    mx = max(arr[i], mx);
  }
  arr[0] = -100000000;
  sort(arr + 1, arr + n + 1);
  int pre = 0;
  dp[0][0] = 1;
  int ans = 0;
  for (int d = 1; arr[1] + d * (k - 1) <= mx; d++) {
    memset(dp, 0, sizeof dp);
    dp[0][0] = 1;
    for (int j = 1; j <= k; j++) {
      int sum = 0;
      int l = 0;
      for (int i = 1; i <= n; i++) {
        while (l < i && arr[l] + d <= arr[i]) {
          sum = (sum + dp[l][(j + 1) % 2]) % 998244353;
          l++;
        }
        dp[i][j % 2] = sum;
      }
      dp[0][0] = 0;
    }
    int cur = 0;
    for (int i = 1; i <= n; i++) {
      cur = (cur + dp[i][k % 2]) % 998244353;
    }
    ans += cur;
    if (ans >= 998244353) ans -= 998244353;
  }
  cout << ans << '\n';
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
