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>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
using namespace std;
const int maxn = 1000 + 5;
const int mod = 998244353;
int dp[maxn][maxn], a[maxn], ans, n, k;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  sort(a + 1, a + n + 1);
  auto solve = [&](int x) {
    dp[0][0] = 1;
    int tmp = 0;
    for (int i = 1; i <= n; i++) {
      while (tmp + 1 <= i && a[i] - a[tmp + 1] >= x) ++tmp;
      for (int j = 0; j <= k; j++) dp[i][j] = dp[i - 1][j];
      for (int j = 1; j <= k; j++) dp[i][j] = (dp[i][j] + dp[tmp][j - 1]) % mod;
    }
    return dp[n][k];
  };
  for (int i = 1; i <= (1e5 / (k - 1)); i++) ans = (ans + solve(i)) % mod;
  printf("%d\n", ans);
  return 0;
}
