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 long long mod = 998244353ll;
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  long long ans = 0;
  for (long long x = 1; x <= (a.back() + k - 2ll) / (k - 1); x++) {
    vector<long long> dp(n);
    vector<long long> pref(n, 1);
    for (long long i = 0; i < n; i++) {
      pref[i] = i + 1;
    }
    long long stop;
    for (long long cnt = 2; cnt <= k; cnt++) {
      stop = 0;
      dp.assign(n, 0);
      for (long long last = cnt - 1; last < n; last++) {
        while (stop < last && a[stop] <= a[last] - x) stop++;
        if (stop != 0) dp[last] = pref[stop - 1];
        dp[last] %= mod;
        if (cnt == k) ans += dp[last], ans %= mod;
      }
      pref[0] = dp[0];
      pref[0] %= mod;
      for (long long i = 1; i < n; i++) {
        pref[i] = pref[i - 1] + dp[i];
        pref[i] %= mod;
      }
    }
  }
  cout << ans << '\n';
}
