#include <bits/stdc++.h>
using namespace std;
const int Mod = 998244353;
int ans, a[1005], n, k, f[1005][1005], sum[1005];
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + 1 + n);
  int lim = 100000 / (k - 1);
  for (int c = 1; c <= lim; c++) {
    memset(sum, 0, sizeof sum);
    int pt = 0;
    for (int i = 1; i <= n; i++) {
      while (a[i] - a[pt + 1] >= c) {
        pt++;
        for (int j = 1; j <= k; j++) sum[j] = (sum[j] + f[pt][j]) % Mod;
      }
      f[i][1] = 1;
      for (int j = 2; j <= k; j++) f[i][j] = sum[j - 1];
    }
    for (int i = 1; i <= n; i++) ans = (ans + f[i][k]) % Mod;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Develop a solution in CPP 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;
const int Mod = 998244353;
int ans, a[1005], n, k, f[1005][1005], sum[1005];
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + 1 + n);
  int lim = 100000 / (k - 1);
  for (int c = 1; c <= lim; c++) {
    memset(sum, 0, sizeof sum);
    int pt = 0;
    for (int i = 1; i <= n; i++) {
      while (a[i] - a[pt + 1] >= c) {
        pt++;
        for (int j = 1; j <= k; j++) sum[j] = (sum[j] + f[pt][j]) % Mod;
      }
      f[i][1] = 1;
      for (int j = 2; j <= k; j++) f[i][j] = sum[j - 1];
    }
    for (int i = 1; i <= n; i++) ans = (ans + f[i][k]) % Mod;
  }
  cout << ans << endl;
  return 0;
}

```