#include <bits/stdc++.h>
using namespace std;
long long mod = 998244353;
int n, k;
int t[1001];
long long dp[1001][1001];
long long kom[1001][1001];
long long sum;
long long solve(int a) {
  for (int i = 1; i <= n; i++) {
    dp[i][1] = 1;
    kom[i][1] = i;
  }
  int x = 0;
  for (int i = 1; i <= n; i++) {
    while (t[i] - t[x + 1] >= a) {
      x++;
    }
    for (int j = 2; j <= k; j++) {
      dp[i][j] = kom[x][j - 1];
      kom[i][j] = kom[i - 1][j] + dp[i][j];
      kom[i][j] %= mod;
    }
  }
  return kom[n][k];
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> t[i];
  }
  sort(t + 1, t + n + 1);
  for (int i = 1; i <= 100000 / (k - 1); i++) {
    sum += solve(i);
  }
  cout << sum % mod;
  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;
long long mod = 998244353;
int n, k;
int t[1001];
long long dp[1001][1001];
long long kom[1001][1001];
long long sum;
long long solve(int a) {
  for (int i = 1; i <= n; i++) {
    dp[i][1] = 1;
    kom[i][1] = i;
  }
  int x = 0;
  for (int i = 1; i <= n; i++) {
    while (t[i] - t[x + 1] >= a) {
      x++;
    }
    for (int j = 2; j <= k; j++) {
      dp[i][j] = kom[x][j - 1];
      kom[i][j] = kom[i - 1][j] + dp[i][j];
      kom[i][j] %= mod;
    }
  }
  return kom[n][k];
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> t[i];
  }
  sort(t + 1, t + n + 1);
  for (int i = 1; i <= 100000 / (k - 1); i++) {
    sum += solve(i);
  }
  cout << sum % mod;
  return 0;
}

```