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 double PI = atan(1.0) * 4;
const long long int MOD = 998244353;
long long int dp[1005][1005];
long long int pref_sum[1005][1005];
long long int a[1005] = {0};
int n, k;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  cin >> n >> k;
  for (int i = 0; i < 1005; i++)
    for (int j = 0; j < 1005; j++) {
      dp[i][j] = 0;
      pref_sum[i][j] = 0;
    }
  for (int i = 1; i < n + 1; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  int lim = (100000 / (k - 1));
  long long int ans = 0;
  for (int i = 1; i < n + 1; i++) pref_sum[1][i] = i;
  pref_sum[1][1] = 1;
  for (int x = 1; x < lim + 1; x++) {
    for (int i = 2; i < k + 1; i++) {
      int ptr = 1;
      for (int j = 1; j < n + 1; j++) {
        if (a[j] - a[1] < i) {
          dp[i][j] = 0;
          pref_sum[i][j] = pref_sum[i][j - 1];
        }
        while (a[j] - a[ptr] >= x) ptr++;
        dp[i][j] = pref_sum[i - 1][ptr - 1];
        pref_sum[i][j] = pref_sum[i][j - 1] + dp[i][j];
        if (pref_sum[i][j] >= MOD) pref_sum[i][j] -= MOD;
      }
    }
    for (int i = 1; i < n + 1; i++) {
      ans = ans + dp[k][i];
      if (ans >= MOD) ans -= MOD;
    }
  }
  cout << ans << endl;
  return 0;
}
