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 N = 1e3 + 1, SN = 1e2, lg = 20, sq = 550, inf = 1e18,
                mod = 998244353;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
long long d[N][N];
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, k, ans = 0;
  cin >> n >> k;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  sort((a).begin(), (a).end());
  for (long long u = 1; u < 2e5; u++) {
    if (u * k > 2e5) break;
    for (long long i = 0; i < n; i++) d[i][1] = 1;
    for (long long c = 2; c <= k; c++) {
      long long sum = 0, it = 0;
      for (long long i = 0; i < n; i++) {
        while (it < i && a[i] - a[it] >= u) {
          sum += d[it][c - 1];
          sum %= mod;
          it++;
        }
        d[i][c] = sum;
      }
    }
    for (long long i = 0; i < n; i++) ans = (ans + d[i][k]) % mod;
  }
  cout << ans;
}
