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;
constexpr int64_t kMod = 998244353;
int Calculate(const vector<int>& a, int k, int x) {
  vector<int> ans(a.size(), 1);
  for (int len = 2; len <= k; ++len) {
    int l = 0;
    int sum = 0;
    vector<int> new_ans(a.size(), 0);
    new_ans[0] = 0;
    for (int r = 1; r < a.size(); ++r) {
      while (a[r] >= a[l] + x) {
        sum = (sum + ans[l]) % kMod;
        ++l;
      }
      new_ans[r] = sum;
    }
    ans = new_ans;
  }
  int res = 0;
  for (int i = 0; i < a.size(); ++i) {
    res = (res + ans[i]) % kMod;
  }
  return res;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  for (auto&& e : a) {
    cin >> e;
  }
  sort(begin(a), end(a));
  int res = 0;
  for (int x = 1; x <= (10 + a.back()) / (k - 1); ++x) {
    res = (res + Calculate(a, k, x)) % kMod;
  }
  cout << res << endl;
  return 0;
}
