#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9;
const int N = 1e3 + 100;
const long long mod = 998244353;
int n, k, a[N];
long long f[N][N], res[N], ans, s[N][N];
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1);
  a[0] = -INF;
  for (int v = 1; v <= a[n] / (k - 1); v++) {
    f[0][0] = 1;
    int now = 0;
    for (int i = 1; i <= n; i++) {
      while (a[i] - a[now + 1] >= v) now++;
      f[i][0] = f[i - 1][0];
      for (int j = 1; j <= k; j++)
        f[i][j] = (f[i - 1][j] + f[now][j - 1]) % mod;
    }
    ans = (ans + f[n][k]) % mod;
  }
  cout << ans << endl;
}

### 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 INF = 1e9;
const int N = 1e3 + 100;
const long long mod = 998244353;
int n, k, a[N];
long long f[N][N], res[N], ans, s[N][N];
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1);
  a[0] = -INF;
  for (int v = 1; v <= a[n] / (k - 1); v++) {
    f[0][0] = 1;
    int now = 0;
    for (int i = 1; i <= n; i++) {
      while (a[i] - a[now + 1] >= v) now++;
      f[i][0] = f[i - 1][0];
      for (int j = 1; j <= k; j++)
        f[i][j] = (f[i - 1][j] + f[now][j - 1]) % mod;
    }
    ans = (ans + f[n][k]) % mod;
  }
  cout << ans << endl;
}

```