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>
const int maxn = 1010;
const int mod = 998244353;
inline void reduce(int& x) { x += x >> 31 & mod; }
struct pr {
  int a, b;
};
int n, k;
int a[maxn];
int f[maxn][maxn];
int pre[maxn][maxn];
int main() {
  std::ios::sync_with_stdio(false), std::cin.tie(0);
  std::cin >> n >> k;
  for (int i = 1; i <= n; ++i) std::cin >> a[i];
  std::sort(a + 1, a + n + 1);
  int ans = 0;
  for (int i = 0; i <= 100000 / (k - 1); ++i) {
    for (int j = 0; j <= n; ++j) {
      std::fill(f[j], f[j] + k + 1, 0);
      std::fill(pre[j], pre[j] + k + 1, 0);
    }
    **pre = 1;
    int p = 1;
    for (int j = 1; j <= n; ++j) {
      while (p < j && a[j] - a[p] > i) ++p;
      for (int l = 0; l <= k; ++l) {
        if (l != 0) {
          f[j][l] = pre[p - 1][l - 1];
        }
        reduce(pre[j][l] = pre[j - 1][l] + f[j][l] - mod);
      }
      reduce(ans += f[j][k] - mod);
    }
  }
  std::cout << ans << '\n';
}
