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;
void add(int& x, int y) {
  x += y;
  if (x >= 998244353) x -= 998244353;
}
int f[1010][1010], g[1010][1010];
int n, k;
int a[1010];
int solve(int dis) {
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= k; j++) f[i][j] = g[i][j] = 0;
  int mx = 0;
  for (int i = 1; i <= n; i++) {
    f[i][1] = 1;
    while (a[i] - a[mx + 1] >= dis) mx++;
    for (int j = 2; j <= k; j++) f[i][j] = g[mx][j - 1];
    for (int j = 1; j <= k; j++) {
      g[i][j] = g[i - 1][j];
      add(g[i][j], f[i][j]);
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) add(ans, f[i][k]);
  return ans;
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1);
  int N = 100000;
  int ans = 0;
  for (int i = 1; i <= N / (k - 1) + 5; i++) add(ans, solve(i));
  printf("%d\n", ans);
}
