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;
long long read() {
  long long k = 1, x = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') k = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + (c - 48);
    c = getchar();
  }
  return x * k;
}
long long n, m, f[2001][2001], s[2001][2001], res[300000], p = 998244353;
long long a[1000604], k, ans, now;
int main() {
  n = read();
  k = read();
  for (int i = 1; i <= n; i++) {
    a[i] = read();
  }
  ans = 0;
  sort(a + 1, a + 1 + n);
  for (int v = 1; v * (k - 1) <= a[n]; v++) {
    now = 0;
    f[0][0] = 1;
    for (int i = 1; i <= n; i++) {
      while (a[i] - a[now + 1] >= v) now++;
      f[i][0] = f[i - 1][0];
      for (int j = 1; j <= k; j++) {
        f[i][j] = (f[i - 1][j] + f[now][j - 1]) % p;
      }
    }
    ans = (ans + f[n][k]) % p;
  }
  printf("%lld\n", ans % p);
}
