#include <bits/stdc++.h>
using namespace std;
const int Maxn = 1000;
const int Mod = 998244353;
int N, K, A[Maxn + 5];
long long f[Maxn + 5][Maxn + 5];
int Solve(int x) {
  f[0][0] = 1;
  int last = 0;
  for (int i = 1; i <= N; i++) {
    while (last + 1 <= i && A[i] - A[last + 1] >= x) last++;
    for (int j = 0; j <= K; j++) f[i][j] = f[i - 1][j];
    for (int j = 1; j <= K; j++) f[i][j] = (f[i][j] + f[last][j - 1]) % Mod;
  }
  return f[N][K];
}
int main() {
  scanf("%d %d", &N, &K);
  for (int i = 1; i <= N; i++) scanf("%d", &A[i]);
  sort(A + 1, A + N + 1);
  long long ans = 0;
  for (int i = 1; i * (K - 1) <= A[N]; i++) ans = (ans + Solve(i)) % Mod;
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please create a solution in CPP 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 = 1000;
const int Mod = 998244353;
int N, K, A[Maxn + 5];
long long f[Maxn + 5][Maxn + 5];
int Solve(int x) {
  f[0][0] = 1;
  int last = 0;
  for (int i = 1; i <= N; i++) {
    while (last + 1 <= i && A[i] - A[last + 1] >= x) last++;
    for (int j = 0; j <= K; j++) f[i][j] = f[i - 1][j];
    for (int j = 1; j <= K; j++) f[i][j] = (f[i][j] + f[last][j - 1]) % Mod;
  }
  return f[N][K];
}
int main() {
  scanf("%d %d", &N, &K);
  for (int i = 1; i <= N; i++) scanf("%d", &A[i]);
  sort(A + 1, A + N + 1);
  long long ans = 0;
  for (int i = 1; i * (K - 1) <= A[N]; i++) ans = (ans + Solve(i)) % Mod;
  printf("%lld\n", ans);
  return 0;
}

```