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 a, b, d[1001], y[100001], dp[1001][1001];
int main() {
  scanf("%lld%lld", &a, &b);
  for (int i = 1; i <= a; i++) {
    scanf("%lld", &d[i]);
  }
  sort(d + 1, d + a + 1);
  long long pp = 1;
  for (int i = d[1]; i <= 100000; i++) {
    while (pp < a && i >= d[pp + 1]) {
      pp++;
    }
    y[i] = pp;
  }
  long long ans = 0;
  for (int i = 0; i <= a; i++) {
    dp[i][0] = 1;
  }
  for (int ii = 1; ii * (b - 1) <= d[a]; ii++) {
    for (int i = 1; i <= a; i++) {
      for (int j = 1; j <= b; j++) {
        dp[i][j] = dp[i - 1][j] + dp[d[i] - ii >= 0 ? y[d[i] - ii] : 0][j - 1];
        dp[i][j] %= 998244353;
      }
    }
    ans += dp[a][b];
    ans %= 998244353;
  }
  printf("%lld", ans);
  return 0;
}
