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>
#pragma GCC optimize("O3")
using namespace std;
int n, k;
int v[1002];
int dp[1002][1002];
int ans[100002];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) cin >> v[i];
  sort(v + 1, v + n + 1);
  long long sol = 0;
  int sum = 0;
  for (int p = 1; p <= n; ++p) dp[p][1] = p;
  for (int i = 100000 / (k - 1); i >= 1; --i) {
    for (int j = 2; j <= k; ++j) {
      int lp = 1;
      for (int p = 1; p <= n; ++p) {
        while (v[p] - v[lp] >= i) ++lp;
        dp[p][j] = dp[lp - 1][j - 1];
        dp[p][j] += dp[p - 1][j];
        if (dp[p][j] >= 998244353) dp[p][j] -= 998244353;
      }
    }
    ans[i] = (dp[n][k] - sum + 998244353);
    if (ans[i] >= 998244353) ans[i] -= 998244353;
    sol += 1LL * i * ans[i];
    sol %= 998244353;
    sum += ans[i];
    if (sum >= 998244353) sum -= 998244353;
  }
  cout << sol << '\n';
  return 0;
}
