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;
const int mod = 998244353;
inline int mul(int x, int y) { return (long long)x * y % mod; }
inline void add(int &x, int y) {
  y += x;
  x = y >= mod ? y - mod : y;
}
inline int inc(int x, int y) {
  x += y;
  return x >= mod ? x - mod : x;
}
const int maxn = 1005;
int n, k, a[maxn], f[maxn][maxn];
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + 1 + n);
  int lim = (a[n] - a[1]) / (k - 1);
  int ans = 0;
  f[0][0] = 1;
  for (int l = 1; l <= lim; l++) {
    for (int i = 1, p = 0; i <= n; i++) {
      while (p < i - 1 && a[i] - a[p + 1] >= l) p++;
      f[i][0] = f[i - 1][0];
      for (int j = 1; j <= min(i, k); j++)
        f[i][j] = inc(f[i - 1][j], f[p][j - 1]);
    }
    add(ans, f[n][k]);
  }
  cout << ans << endl;
  return 0;
}
