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;
using ll = long long;
const int SIZE = 1e5;
const int MOD = 998244353;
void add(int &lhs, int rhs) {
  lhs = lhs + rhs;
  if (lhs >= MOD) lhs -= MOD;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a.begin(), a.end());
  vector<int> ans(SIZE + 2);
  vector<vector<int>> dp(n + 1, vector<int>(k + 1));
  for (int d = 0; d <= SIZE; d++) {
    if (d * (k - 1) > SIZE) continue;
    for (int i = 0; i <= n; i++) fill(dp[i].begin(), dp[i].end(), 0);
    dp[0][0] = 1;
    int r = 1;
    for (int i = 0; i < n; i++) {
      while (r < n && (i == r || a[r] - a[i] < d)) r++;
      for (int j = 0; j < k; j++) {
        add(dp[i + 1][j], dp[i][j]);
        add(dp[r][j + 1], dp[i][j]);
      }
      add(dp[i + 1][k], dp[i][k]);
    }
    ans[d] = dp[n][k];
  }
  int res = 0;
  for (int i = 0; i <= SIZE; i++) {
    int cnt = ((ans[i] - ans[i + 1]) % MOD + MOD) % MOD;
    add(res, ll(cnt) * i % MOD);
  }
  cout << res << endl;
  return 0;
}
