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;
using VL = vector<ll>;
using VI = vector<int>;
const ll MOD = 998244353;
inline void suma(ll& x, ll y) { x = (x + y) % MOD; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k;
  cin >> n >> k;
  VI A(n);
  for (int& a : A) cin >> a;
  A.push_back(-1e9);
  sort(A.begin(), A.end());
  ll res = 0;
  int mx = A[n] / (k - 1);
  VI ini(n + 1, n);
  for (int b = 1; b <= mx; ++b) {
    for (int j = 1; j <= n; ++j)
      while (A[ini[j]] + b > A[j]) --ini[j];
    VL pre1(n + 1, 0);
    for (int j = 1; j <= n; ++j) pre1[j] = j;
    for (int i = 2; i <= k; ++i) {
      VL pre2(n + 1, 0);
      for (int j = i; j <= n; ++j) suma(pre2[j], pre2[j - 1] + pre1[ini[j]]);
      swap(pre1, pre2);
    }
    suma(res, pre1[n]);
  }
  cout << res << endl;
}
