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.
#include <bits/stdc++.h>
using namespace std;
long long int arr[1001];
long long int ans[100001];
long long int dp[1001][1001];
long long int c[1001];
bool check(long long int a, long long int b) { return a > b; }
int main() {
  long long int n, k;
  cin >> n >> k;
  long long int maxi = 0;
  for (long long int i = 1; i <= n; i++) {
    cin >> arr[i];
    maxi = max(maxi, arr[i]);
  }
  sort(arr + 1, arr + (n + 1), check);
  for (long long int j = 1; j <= maxi; j++) {
    ans[j] = 0;
  }
  for (long long int j = 1; j <= (long long int)(maxi / (k - 1)) + 2; j++) {
    long long int t = 0;
    for (long long int i = 0; i <= n; i++) {
      for (long long int v = 0; v <= k; v++) {
        dp[i][v] = 0;
      }
    }
    for (long long int i = 1; i <= n; i++) {
      dp[i][1] = i;
      while (arr[t + 1] >= arr[i] + j && t < n) {
        t++;
      }
      for (long long int v = 2; v <= k; v++) {
        dp[i][v] = dp[t][v - 1];
        dp[i][v] = (dp[i][v] + dp[i - 1][v]) % 998244353;
      }
    }
    if (j == 1) {
    }
    ans[j] = dp[n][k];
  }
  long long int res = 0;
  for (long long int i = 1; i <= maxi; i++) {
    res = (res + ans[i]) % 998244353;
  }
  cout << res << "\n";
}
