#include <bits/stdc++.h>
using namespace std;
const int p = 998244353;
const int N = 1005;
int a[N], f[N][N], s[N * 100];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1);
  int lim = (a[n] - a[1]) / (k - 1);
  for (int v = 1; v <= lim; v++) {
    for (int i = 0; i <= n; i++)
      for (int j = 1; j <= k; j++) f[i][j] = 0;
    for (int i = 1, j = 0; i <= n; i++) {
      while (j + 1 < i && a[i] - a[j + 1] >= v) j++;
      f[i][1] = f[i - 1][1] + 1;
      for (int t = 2; t <= k; t++) f[i][t] = (f[j][t - 1] + f[i - 1][t]) % p;
    }
    s[v] = f[n][k];
  }
  int ans = 0;
  for (int v = 1; v <= lim; v++)
    (ans += (long long)v * (s[v] - s[v + 1] + p) % p) %= p;
  printf("%d\n", ans);
  return 0;
}

### Prompt

Your challenge is to write 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 p = 998244353;
const int N = 1005;
int a[N], f[N][N], s[N * 100];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1);
  int lim = (a[n] - a[1]) / (k - 1);
  for (int v = 1; v <= lim; v++) {
    for (int i = 0; i <= n; i++)
      for (int j = 1; j <= k; j++) f[i][j] = 0;
    for (int i = 1, j = 0; i <= n; i++) {
      while (j + 1 < i && a[i] - a[j + 1] >= v) j++;
      f[i][1] = f[i - 1][1] + 1;
      for (int t = 2; t <= k; t++) f[i][t] = (f[j][t - 1] + f[i - 1][t]) % p;
    }
    s[v] = f[n][k];
  }
  int ans = 0;
  for (int v = 1; v <= lim; v++)
    (ans += (long long)v * (s[v] - s[v + 1] + p) % p) %= p;
  printf("%d\n", ans);
  return 0;
}

```