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 N = 1e3 + 5;
const int MOD = 998244353;
int n, k, a[N], p[N], f[N][N], sum[N][N];
void add(int &x, int y) { (x += y) >= MOD && (x -= MOD); }
int solve(int x) {
  for (int i = 0; i <= k; i++) {
    for (int j = 0; j <= n; j++) {
      f[i][j] = sum[i][j] = 0;
    }
  }
  for (int i = 1, j = 0; i <= n; i++) {
    for (; a[i] - a[j + 1] >= x; j++)
      ;
    p[i] = j;
  }
  f[0][0] = 1;
  std::fill(sum[0], sum[0] + n + 1, 1);
  for (int i = 1; i <= k; i++) {
    for (int j = i; j <= n; j++) {
      f[i][j] = sum[i - 1][p[j]];
      sum[i][j] = f[i][j];
    }
    for (int j = i; j <= n; j++) {
      add(sum[i][j], sum[i][j - 1]);
    }
  }
  return sum[k][n];
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  std::sort(a + 1, a + n + 1);
  int ans = 0;
  for (int t = a[n], i = 1; i * (k - 1) <= t; i++) add(ans, solve(i));
  printf("%d\n", ans);
  return 0;
}
