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>
#pragma comment(linker, "/STACK:256000000")
#pragma GCC optimize("O3")
using namespace std;
const int maxN = 1e3 + 2, mod = 998244353;
int n, k;
int a[maxN];
long long dp[maxN][maxN], pr[maxN][maxN], max_ind[maxN];
long long solve(int val) {
  int j = 0;
  for (int i = 0; i < int(n); i++) {
    while ((a[i] - a[j]) > val) j++;
    max_ind[i] = j - 1;
  }
  for (int i = 0; i < int(n); i++) {
    dp[1][i] = 1;
    pr[1][i] = i + 1;
  }
  for (int len = 2; len <= k; len++) pr[len][len - 2] = 0;
  for (int len = 2; len <= k; len++)
    for (int ind = len - 1; ind < n; ind++) {
      if (max_ind[ind] < 0)
        dp[len][ind] = 0;
      else
        dp[len][ind] = pr[len - 1][max_ind[ind]];
      pr[len][ind] = (pr[len][ind - 1] + dp[len][ind]) % mod;
    }
  long long ans = 0;
  for (int ind = k - 1; ind < n; ind++) {
    ans += dp[k][ind];
    ans %= mod;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 0; i < int(n); i++) cin >> a[i];
  sort(a, a + n);
  int maxD = a[n - 1] / (k - 1) + 1;
  long long ans = 0;
  for (int i = 0; i < int(maxD); i++) {
    ans += solve(i);
    ans %= mod;
  }
  cout << ans;
  return 0;
}
