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;
const long long mod = 998244353;
const int ms = 1e3 + 20;
long long dp[ms][ms];
int a[ms];
int n;
long long solve(int x, int mk) {
  for (int i = 0; i < n; i++) {
    dp[0][i] = 1;
  }
  for (int i = 1; i < mk; i++) {
    dp[i][n - 1] = 0;
  }
  for (int k = 1; k < mk; k++) {
    long long ans = 0;
    int j = n - 1;
    for (int i = n - 1; i >= 0; i--) {
      while (j > i && a[j] - a[i] >= x) {
        ans += dp[k - 1][j];
        j--;
      }
      ans %= mod;
      dp[k][i] = ans;
    }
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    ans += dp[mk - 1][i];
  }
  return ans % mod;
}
int main() {
  int k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    scanf(" %d", a + i);
  }
  sort(a, a + n);
  long long ans = 0;
  for (int x = 1; a[0] + (k - 1) * x <= a[n - 1]; x++) {
    ans += solve(x, k);
  }
  cout << ans % mod << endl;
  return 0;
}
