#include <bits/stdc++.h>
using namespace std;
long long n, k;
long long dp[1002][1002];
long long a[1002];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(), cout.tie();
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  long long ans = 0;
  for (int i = 1; i <= (a[n] - a[1]) / (k - 1); i++) {
    long long l = 0;
    for (int j = 1; j <= n; j++) {
      while (l < j && a[j] - a[l + 1] >= i) l++;
      for (int x = 1; x <= k; x++) {
        dp[j][x] = dp[j - 1][x] + dp[l][x - 1] + (x == 1);
        if (dp[j][x] >= 998244353) dp[j][x] -= 998244353;
      }
    }
    ans = (ans + dp[n][k]) % 998244353;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long n, k;
long long dp[1002][1002];
long long a[1002];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(), cout.tie();
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  long long ans = 0;
  for (int i = 1; i <= (a[n] - a[1]) / (k - 1); i++) {
    long long l = 0;
    for (int j = 1; j <= n; j++) {
      while (l < j && a[j] - a[l + 1] >= i) l++;
      for (int x = 1; x <= k; x++) {
        dp[j][x] = dp[j - 1][x] + dp[l][x - 1] + (x == 1);
        if (dp[j][x] >= 998244353) dp[j][x] -= 998244353;
      }
    }
    ans = (ans + dp[n][k]) % 998244353;
  }
  cout << ans << endl;
  return 0;
}

```