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 my_upper_bound(int a[], int l, int r, int val) {
  while (l != r) {
    int m = l + (r - l) / 2;
    if (a[m] <= val) {
      l = m + 1;
    } else {
      r = m;
    }
  }
  return l;
}
const int n0 = 1003, m = 1e5, mod = 998244353;
int n, k, a[n0], dp[2][n0], b[n0];
int calc(int gap) {
  for (int i = 1; i <= n; i++)
    b[i] = my_upper_bound(a, 1, n + 1, a[i] - gap) - 1;
  memset(&dp, 0, sizeof(dp));
  for (int j = 0; j <= n; j++) dp[0][j] = 1;
  for (int i = 1; i <= k; i++) {
    dp[i & 1][0] = 0;
    for (int j = 1; j <= n; j++) {
      int &x = dp[i & 1][j];
      x = dp[i & 1][j - 1] + dp[(i - 1) & 1][b[j]];
      x = (x >= mod) ? (x - mod) : x;
    }
  }
  return dp[k & 1][n];
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int karan = 100, great = 30;
  while (karan--) great = great % 4 + 3;
  if (great == 7) return 0;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  int res = 0;
  for (int gap = 1; gap <= m / (k - 1); gap++) {
    res += calc(gap);
    if (res >= mod) res -= mod;
  }
  cout << res;
}
