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;
int a[1050], dp[1050][1050];
inline void read(int &);
int main() {
  int n, K, lim(0);
  read(n), read(K);
  for (int i(1); i <= n; ++i) read(a[i]), lim = max(lim, a[i]);
  sort(a + 1, a + n + 1), a[0] = -0x3f3f3f3f;
  int i, j, k, l, r, ans(0);
  for (i = 1; i <= lim / (K - 1); ++i) {
    for (j = 1, dp[0][0] = 1; j <= K; ++j)
      for (k = j, l = j - 1, r = 0; k <= n; ++k) {
        while (a[k] - a[l] >= i) r = (r + dp[j - 1][l]) % 998244353, ++l;
        dp[j][k] = r;
      }
    for (k = K; k <= n; ++k) ans = (ans + dp[K][k]) % 998244353;
  }
  printf("%d", ans);
  return 0;
}
inline void read(int &x) {
  x ^= x;
  register char c;
  while (c = getchar(), c < '0' || c > '9')
    ;
  while (c >= '0' && c <= '9')
    x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
}
