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, mod = 998244353;
vector<int> a;
int solve(int x) {
  if (x > a.back() / (k - 1)) return 0;
  vector<int> dp(n + 1, 1);
  dp.back() = 0;
  for (int itt = 2; itt <= k; itt++) {
    for (int i = n - 1; i >= 0; i--) {
      dp[i] += dp[i + 1];
      if (dp[i] >= mod) dp[i] -= mod;
    }
    vector<int> new_dp(n + 1);
    int pt = n;
    for (int i = n - 1; i >= 0; i--) {
      while (a[i] + x <= a[pt - 1]) pt--;
      new_dp[i] = dp[pt];
    }
    dp = new_dp;
  }
  int ret = 0;
  for (auto &i : dp) {
    ret += i;
    if (ret >= mod) ret -= mod;
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  a.resize(n);
  for (auto &i : a) cin >> i;
  sort(a.begin(), a.end());
  int ans = 0;
  for (int i = 1; i <= a.back(); i++) {
    ans += solve(i);
    if (ans >= mod) ans -= mod;
  }
  cout << ans << endl;
}
