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;
const int N = 5010;
int n, k, a[N], f[N][N], s[N][N], res[100010], 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);
  a[0] = -1e9;
  for (int i = 1; i * (k - 1) <= a[n]; i++) {
    f[0][0] = 1;
    int now = 0;
    s[0][0] = 1;
    for (int j = 1; j <= n; j++) {
      while (a[now] <= a[j] - i) now++;
      for (int m = 0; m <= k; m++) {
        if (m) f[j][m] = s[now - 1][m - 1];
        s[j][m] = (s[j - 1][m] + f[j][m]) % mod;
      }
      res[i] = (res[i] + f[j][k]) % mod;
    }
    ans = (ans + res[i]) % mod;
  }
  printf("%d", ans);
}
