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:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
int n, k;
long long a[1000], cnt[100000 + 2];
long long dp[1000][1000 + 1];
long long solve(int minstep) {
  int idx = 0;
  for (int i = 0; i < (n); i++) {
    for (int j = 0; j < (k + 1); j++) dp[i][j] = 0;
    dp[i][1] = 1;
    while (a[idx] <= a[i] - minstep) idx++;
    if (i)
      for (int j = (1); j < (k + 1); j++)
        dp[i][j] = (dp[i][j] + dp[i - 1][j]) % 998244353;
    if (idx)
      for (int j = (1); j < (k + 1); j++)
        dp[i][j] = (dp[i][j] + dp[idx - 1][j - 1]) % 998244353;
  }
  return dp[n - 1][k];
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < (n); i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  for (int i = (1); i < (100000 / (k - 1) + 1); i++) cnt[i] = solve(i);
  long long res = 0;
  for (int i = (1); i < (100000 / (k - 1) + 1); i++) {
    res += i * ((cnt[i] + 998244353 - cnt[i + 1]) % 998244353) % 998244353;
    res %= 998244353;
  }
  cout << res << "\n";
}
