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;
long long date[100010], dp[1001][1001], mod = 998244353;
int main() {
  long long n, k, sol = 0;
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) cin >> date[i];
  sort(date + 1, date + n + 1);
  long long h = 0;
  for (long long x = date[n] / (k - 1); x >= 1; x--) {
    for (long long i = 1; i <= k; i++)
      for (long long j = 1; j <= n; j++) dp[i][j] = 0;
    for (long long i = 1; i <= n; i++) dp[1][i] = 1;
    long long sum = 0;
    for (long long i = 2; i <= k; i++) {
      sum = 0;
      long long p = i - 1;
      for (long long j = i; j <= n; j++) {
        while (date[j] - date[p] >= x) sum += dp[i - 1][p], p++, sum %= mod;
        dp[i][j] = sum;
      }
    }
    if (x == 9) {
      sum = 0;
    }
    sum = 0;
    for (long long i = k; i <= n; i++) sum = sum + dp[k][i], sum %= mod;
    long long y = sum - h;
    if (y < 0) y += mod;
    sol = sol + y * x % mod;
    sol %= mod;
    h = sum;
  }
  cout << sol;
  return 0;
}
