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;
template <typename A, typename B = A>
using pr = pair<A, B>;
const int mod = 998244353;
int main() {
  ios::sync_with_stdio(false), cin.tie(nullptr);
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  for (auto &x : a) cin >> x;
  sort((a).begin(), (a).end());
  int sz = *max_element((a).begin(), (a).end()) / (k - 1) + 2;
  vector<int> dp1(sz);
  vector<vector<int>> pref(n + 1, vector<int>(k + 1));
  vector<vector<int>> dp(n + 1, vector<int>(k + 1));
  a.insert(a.begin(), 0);
  for (int delta = 0; delta < sz; delta++) {
    dp[0][0] = 1;
    pref[0][0] = 1;
    int prev = 1;
    for (int i = 1; i <= n; i++) {
      while (prev < i && a[i] - a[prev] >= delta) prev++;
      for (int j = 0; j <= k; j++) {
        if (j) dp[i][j] = pref[prev - 1][j - 1];
        pref[i][j] = pref[i - 1][j] + dp[i][j];
        if (pref[i][j] >= mod) pref[i][j] -= mod;
      }
      dp1[delta] += dp[i][k];
      if (dp1[delta] >= mod) dp1[delta] -= mod;
    }
  }
  int ans = dp1[sz - 1];
  for (int i = sz - 2; i >= 0; i--) {
    int cur = dp1[i] - dp1[i + 1];
    if (cur < 0) cur += mod;
    ans += 1ll * cur * i % mod;
    if (ans >= mod) ans -= mod;
  }
  cout << ans << '\n';
}
