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;
int n, k;
int values[1013];
long long dp[1013][1013];
long long prefix[1013][1013];
int main() {
  if (fopen("FILENAME.in", "r")) {
    freopen("FILENAME.in", "r", stdin);
    freopen("FILENAME.out", "w", stdout);
  }
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> values[i];
  }
  sort(values, values + n);
  long long answer = 0;
  for (int beauty = 1; beauty * (k - 1) <= 100010; beauty++) {
    for (int i = 0; i <= n; i++) {
      for (int j = 0; j <= k; j++) {
        dp[i][j] = 0;
        prefix[i][j] = 0;
      }
    }
    for (int i = 0; i < n; i++) {
      dp[i][1] = 1;
      prefix[i + 1][1] = prefix[i][1] + 1;
    }
    int bound = 0;
    for (int i = 1; i < n; i++) {
      while (abs(values[bound] - values[i]) >= beauty) {
        bound++;
      }
      bound--;
      if (abs(values[0] - values[i]) < beauty) {
        for (int j = 2; j <= k; j++) {
          prefix[i + 1][j] += prefix[i][j];
        }
        continue;
      }
      for (int j = 2; j <= k; j++) {
        dp[i][j] += prefix[bound + 1][j - 1];
        dp[i][j] %= 998244353LL;
        prefix[i + 1][j] += prefix[i][j] + dp[i][j];
        prefix[i + 1][j] %= 998244353LL;
      }
    }
    for (int i = 0; i < n; i++) {
      answer += dp[i][k];
      answer %= 998244353LL;
    }
  }
  cout << answer << endl;
}
