#include <bits/stdc++.h>
using namespace std;
const long long MXN = 1e3 + 10;
const long long MXM = 1e5 + 10;
const long long Mod = 998244353;
long long n, k, ans;
long long A[MXN], dp[MXN][MXN];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> A[i];
  sort(A + 1, A + n + 1);
  for (int d = 1; d * (k - 1) < MXM; d++) {
    dp[0][0] = dp[1][0] = dp[1][1] = 1;
    for (int i = 2; i <= n; i++) {
      dp[i][0] = 1;
      long long p = upper_bound(A + 1, A + i + 1, A[i] - d) - A - 1;
      for (int j = 1; j <= k; j++) {
        dp[i][j] = (dp[i - 1][j] + dp[p][j - 1]) % Mod;
      }
    }
    ans = (ans + dp[n][k]) % Mod;
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

In CPP, your task is to solve the following problem:

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;
const long long MXN = 1e3 + 10;
const long long MXM = 1e5 + 10;
const long long Mod = 998244353;
long long n, k, ans;
long long A[MXN], dp[MXN][MXN];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> A[i];
  sort(A + 1, A + n + 1);
  for (int d = 1; d * (k - 1) < MXM; d++) {
    dp[0][0] = dp[1][0] = dp[1][1] = 1;
    for (int i = 2; i <= n; i++) {
      dp[i][0] = 1;
      long long p = upper_bound(A + 1, A + i + 1, A[i] - d) - A - 1;
      for (int j = 1; j <= k; j++) {
        dp[i][j] = (dp[i - 1][j] + dp[p][j - 1]) % Mod;
      }
    }
    ans = (ans + dp[n][k]) % Mod;
  }
  cout << ans << '\n';
  return 0;
}

```