#include <bits/stdc++.h>
using namespace std;
long long p = 998244353;
long a[1005];
unsigned long s[1005][1005];
int main() {
  long n, k;
  unsigned long sum = 0;
  cin >> n >> k;
  for (long i = 1; i <= n; ++i) {
    scanf("%ld", &a[i]);
  }
  sort(a + 1, a + 1 + n);
  for (long i = 1; i <= k; ++i) s[0][i] = s[1][i] = 0;
  for (long i = 1; i <= n; ++i) s[i][1] = i;
  for (long m = 1; m <= a[n] / (k - 1); ++m) {
    for (long i = 2; i <= n; ++i) {
      long l = upper_bound(a + 1, a + i + 1, a[i] - m) - a;
      for (long j = 2; j <= k; ++j) {
        s[i][j] = (s[l - 1][j - 1] + s[i - 1][j]) % p;
      }
    }
    sum = (sum + s[n][k]) % p;
  }
  cout << sum << endl;
  return 0;
}

### Prompt

Your task is to create a Cpp solution to 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;
long long p = 998244353;
long a[1005];
unsigned long s[1005][1005];
int main() {
  long n, k;
  unsigned long sum = 0;
  cin >> n >> k;
  for (long i = 1; i <= n; ++i) {
    scanf("%ld", &a[i]);
  }
  sort(a + 1, a + 1 + n);
  for (long i = 1; i <= k; ++i) s[0][i] = s[1][i] = 0;
  for (long i = 1; i <= n; ++i) s[i][1] = i;
  for (long m = 1; m <= a[n] / (k - 1); ++m) {
    for (long i = 2; i <= n; ++i) {
      long l = upper_bound(a + 1, a + i + 1, a[i] - m) - a;
      for (long j = 2; j <= k; ++j) {
        s[i][j] = (s[l - 1][j - 1] + s[i - 1][j]) % p;
      }
    }
    sum = (sum + s[n][k]) % p;
  }
  cout << sum << endl;
  return 0;
}

```