#include <bits/stdc++.h>
using namespace std;
long long n, k, arr[1010];
int ptr;
long long dp[1010][1010], pref[1010][1010];
int main() {
  scanf(" %lld%lld", &n, &k);
  long long mv = 0;
  for (int i = 0; i < n; i++) scanf("%lld", &arr[i]), mv = max(mv, arr[i]);
  sort(arr, arr + n);
  long long ans = 0;
  for (int c = 1; c * (k - 1) <= mv; c++) {
    ptr = -1;
    for (int i = 0; i < n; i++)
      for (int j = 1; j <= k; j++) dp[i][j] = 0, pref[i][j] = 0;
    dp[0][0] = 1;
    pref[0][0] = 1;
    dp[0][1] = 1;
    pref[0][1] = 1;
    long long tans = 0;
    for (int i = 1; i < n; i++) {
      dp[i][0] = 1;
      pref[i][0] = 1;
      while (ptr < i && arr[ptr + 1] + c <= arr[i]) ptr++;
      for (int j = 1; j <= k; j++)
        dp[i][j] = (ptr < 0) ? (j == 1) : pref[ptr][j - 1],
        pref[i][j] = (pref[i - 1][j] + dp[i][j] + 998244353) % 998244353;
      tans += dp[i][k];
      tans %= 998244353;
    }
    ans = (ans + tans + 998244353) % 998244353;
  }
  printf("%lld\n", ans);
}

### Prompt

Create a solution in Cpp for 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 n, k, arr[1010];
int ptr;
long long dp[1010][1010], pref[1010][1010];
int main() {
  scanf(" %lld%lld", &n, &k);
  long long mv = 0;
  for (int i = 0; i < n; i++) scanf("%lld", &arr[i]), mv = max(mv, arr[i]);
  sort(arr, arr + n);
  long long ans = 0;
  for (int c = 1; c * (k - 1) <= mv; c++) {
    ptr = -1;
    for (int i = 0; i < n; i++)
      for (int j = 1; j <= k; j++) dp[i][j] = 0, pref[i][j] = 0;
    dp[0][0] = 1;
    pref[0][0] = 1;
    dp[0][1] = 1;
    pref[0][1] = 1;
    long long tans = 0;
    for (int i = 1; i < n; i++) {
      dp[i][0] = 1;
      pref[i][0] = 1;
      while (ptr < i && arr[ptr + 1] + c <= arr[i]) ptr++;
      for (int j = 1; j <= k; j++)
        dp[i][j] = (ptr < 0) ? (j == 1) : pref[ptr][j - 1],
        pref[i][j] = (pref[i - 1][j] + dp[i][j] + 998244353) % 998244353;
      tans += dp[i][k];
      tans %= 998244353;
    }
    ans = (ans + tans + 998244353) % 998244353;
  }
  printf("%lld\n", ans);
}

```