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;
int n, k;
int mod = 998244353;
int a[1005];
int dp[1005][1005];
int c[100005];
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] = -(1e5);
  int ans = 0;
  for (int i = 0; i <= n; ++i) dp[i][0] = 1;
  for (int d = 1; d <= a[n] / (k - 1); ++d) {
    for (int i = 1; i <= n; ++i) memset(dp[i] + 1, 0, k + 1);
    int j = 1;
    for (int i = 1; i <= n; ++i) {
      while (j < i && a[i] - a[j] >= d) ++j;
      for (int len = 1; len <= k; ++len)
        dp[i][len] = (dp[j - 1][len - 1] + dp[i - 1][len]) % mod;
    }
    c[d] = dp[n][k];
  }
  for (int i = 1; i <= a[n]; ++i) {
    int nm = (c[i] - c[i + 1] + mod) % mod;
    int vl = (1LL * nm * i) % mod;
    ans = (ans + vl) % mod;
  }
  printf("%d\n", ans);
  return 0;
}
