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 = 998244353;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  srand(time(NULL));
  long long n, k;
  cin >> n >> k;
  vector<long long> a(n + 1, 0);
  for (long long i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  sort(begin(a), end(a));
  long long Max_value = *prev(end(a)) / (k - 1);
  vector<long long> p(Max_value + 1, 0);
  long long ans = 0;
  vector<vector<long long>> dp(n + 1, vector<long long>(k + 1, 0));
  vector<vector<long long>> pref(n + 1, vector<long long>(k + 1, 0));
  for (long long i = 0; i <= n; ++i) {
    dp[i][0] = 1;
    pref[i][0] = 1;
  }
  for (long long x = 1; x <= Max_value; ++x) {
    long long cur = 0;
    for (long long i = 1; i <= n; ++i) {
      for (long long j = 1; j <= k; ++j) {
        while (a[i] - a[cur + 1] >= x) {
          cur++;
        }
        dp[i][j] = pref[cur][j - 1];
        pref[i][j] = (dp[i][j] + pref[i - 1][j]) % mod;
      }
    }
    for (long long i = 1; i <= n; ++i) {
      p[x] += dp[i][k];
      p[x] %= mod;
    }
    ans += p[x];
  }
  cout << ans % mod;
  return 0;
}
