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>
int read() {
  int ans = 0, c, f = 1;
  while (!isdigit(c = getchar()))
    if (c == '-') f *= -1;
  do ans = ans * 10 + c - '0';
  while (isdigit(c = getchar()));
  return ans * f;
}
const int N = 1050;
const int K = 1050;
const int mod = 998244353;
int n, A[N], f[N][K];
inline void add(int &x, int y) {
  if ((x += y) >= mod) x -= mod;
}
int main() {
  n = read();
  int k = read();
  for (int i = 1; i <= n; ++i) A[i] = read();
  A[0] = -10000000;
  std::sort(A, A + n + 1);
  int ans = 0;
  for (int u = 1; u * (k - 1) <= 100000; ++u) {
    int j = 0;
    for (int i = 1; i <= n; ++i) {
      while (A[j + 1] + u <= A[i]) ++j;
      for (int t = 0; t <= k; ++t) {
        f[i][t] = f[i - 1][t];
        if (t) add(f[i][t], f[j][t - 1]);
      }
      if ((++f[i][1]) >= mod) f[i][1] -= mod;
    }
    add(ans, f[n][k]);
  }
  printf("%d\n", ans);
}
