#include <bits/stdc++.h>
using namespace std;
const int N = 100000, mxn = 1010, md = 998244353;
int n, k, a[mxn], pre[mxn], ans, f[mxn][mxn], q[mxn];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]), pre[i] = i;
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= N / (k - 1); ++i) {
    for (int j = 1; j <= n; ++j)
      for (; pre[j] && a[j] - a[pre[j]] < i; --pre[j])
        ;
    f[0][0] = q[0] = 1;
    for (int j = 1; j <= k; ++j) f[0][j] = q[j] = 0;
    for (int j = 1, cur = 0; j <= n; ++j) {
      for (int l = k; l; --l) f[j][l] = (f[j - 1][l] + f[pre[j]][l - 1]) % md;
      f[j][0] = f[j - 1][0];
    }
    ans = (ans + f[n][k]) % md;
  }
  printf("%d\n", ans);
  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;
const int N = 100000, mxn = 1010, md = 998244353;
int n, k, a[mxn], pre[mxn], ans, f[mxn][mxn], q[mxn];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]), pre[i] = i;
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= N / (k - 1); ++i) {
    for (int j = 1; j <= n; ++j)
      for (; pre[j] && a[j] - a[pre[j]] < i; --pre[j])
        ;
    f[0][0] = q[0] = 1;
    for (int j = 1; j <= k; ++j) f[0][j] = q[j] = 0;
    for (int j = 1, cur = 0; j <= n; ++j) {
      for (int l = k; l; --l) f[j][l] = (f[j - 1][l] + f[pre[j]][l - 1]) % md;
      f[j][0] = f[j - 1][0];
    }
    ans = (ans + f[n][k]) % md;
  }
  printf("%d\n", ans);
  return 0;
}

```