#include <bits/stdc++.h>
using namespace std;
const int maxn = 1005;
const int mod = 998244353;
int f[maxn][maxn], pre[maxn][maxn], n, k, a[maxn], ans;
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + 1 + n);
  for (int i = 1; i <= n; i++) pre[i][1] = i;
  for (int v = 1; v * (k - 1) <= a[n]; v++) {
    int p = 1;
    for (int i = 2; i <= n; i++) {
      while (a[p] + v <= a[i]) p++;
      for (int j = 2; j <= min(i, k); j++) {
        f[i][j] = pre[p - 1][j - 1];
        pre[i][j] = (f[i][j] + pre[i - 1][j]) % mod;
      }
    }
    ans = (ans + pre[n][k]) % mod;
    if (ans == 0) break;
  }
  printf("%d\n", ans);
}

### Prompt

Generate 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;
const int maxn = 1005;
const int mod = 998244353;
int f[maxn][maxn], pre[maxn][maxn], n, k, a[maxn], ans;
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + 1 + n);
  for (int i = 1; i <= n; i++) pre[i][1] = i;
  for (int v = 1; v * (k - 1) <= a[n]; v++) {
    int p = 1;
    for (int i = 2; i <= n; i++) {
      while (a[p] + v <= a[i]) p++;
      for (int j = 2; j <= min(i, k); j++) {
        f[i][j] = pre[p - 1][j - 1];
        pre[i][j] = (f[i][j] + pre[i - 1][j]) % mod;
      }
    }
    ans = (ans + pre[n][k]) % mod;
    if (ans == 0) break;
  }
  printf("%d\n", ans);
}

```