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>
inline long long gi() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) f ^= ch == '-', ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return f ? x : -x;
}
int n, k, a[1010], f[1010][1010];
inline void add(int& a, int b) {
  a = a + b >= 998244353 ? a + b - 998244353 : a + b;
}
inline int solve(int x) {
  for (int i = 1, p = 1; i <= n; ++i) {
    memset(f[i] + 1, 0, 4 * k);
    while (p <= n && a[i] - a[p] >= x) ++p;
    for (int j = 1; j <= k; ++j)
      add(f[i][j], f[i - 1][j]), add(f[i][j], f[p - 1][j - 1] + (j == 1));
  }
  return f[n][k];
}
int main() {
  n = gi(), k = gi();
  for (int i = 1; i <= n; ++i) a[i] = gi();
  std::sort(a + 1, a + n + 1);
  int ans = 0;
  for (int i = 1; i * (k - 1) <= a[n]; ++i) add(ans, solve(i));
  printf("%d\n", ans);
  return 0;
}
