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;
const int MAXN = 1010;
const int MAXK = 1010;
const int MAXB = 100010;
const int MOD = 998244353;
int a[MAXN];
int dp[MAXN][MAXK];
int gb[MAXB];
int psum[MAXN][MAXK];
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  int maxb = min(MAXB - 1, MAXB / (k - 1) + 1);
  for (int b = 0; b <= maxb; b++) {
    for (int i = 0; i < n; i++)
      for (int j = 0; j <= k; j++) {
        dp[i][j] = 0;
        psum[i][j] = 0;
      }
    for (int i = 0; i < n; i++) {
      dp[i][1] = 1;
      psum[i][1] = i + 1;
    }
    int prvi = -1;
    for (int i = 1; i < n; i++) {
      while (prvi < i - 1 && a[prvi + 1] <= a[i] - b) {
        prvi += 1;
      }
      for (int j = 2; j <= k; j++) {
        if (prvi == -1) {
          dp[i][j] = 0;
        } else {
          dp[i][j] = psum[prvi][j - 1];
        }
        psum[i][j] = psum[i - 1][j] + dp[i][j];
        psum[i][j] %= MOD;
      }
    }
    gb[b] = 0;
    for (int i = 0; i < n; i++) {
      gb[b] += dp[i][k];
      gb[b] %= MOD;
    }
  }
  int res = 0;
  for (int b = 0; b < maxb; b++) {
    int t = ((long long)b * ((long long)gb[b] - gb[b + 1])) % MOD + MOD;
    t %= MOD;
    res += t;
    res = res % MOD;
  }
  cout << res << endl;
}
