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;
int n, k;
int a[1015], ans[100015];
int f[1015][1015];
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') f = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
  return x * f;
}
int main() {
  n = read(), k = read();
  int mx = 0;
  for (int i = 1; i <= n; i++) a[i] = read();
  sort(a + 1, a + n + 1);
  for (int res = 0; res <= (100000 / (k - 1)) + 5; res++) {
    for (int i = 0; i <= n; i++)
      for (int j = 0; j <= min(i, k); j++) f[i][j] = 0;
    f[0][0] = 1;
    for (int i = 1, x = 0; i <= n; i++) {
      for (; x + 1 < i && a[i] - a[x + 1] >= res; ++x)
        ;
      f[i][0] = 1;
      for (int j = 1; j <= min(i, k); j++)
        f[i][j] = (f[i - 1][j] + f[x][j - 1]) % mod;
    }
    ans[res] = f[n][k];
  }
  int pps = 0;
  for (int i = 1; i <= (100000 / (k - 1)) + 5; i++)
    pps = (pps + 1ll * i * (ans[i] - ans[i + 1])) % mod;
  pps = (pps + mod) % mod;
  printf("%d\n", pps);
  return 0;
}
