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, w[1010], S[1010][1010], Mod = 998244353;
int Z[101000];
int Calc(int KK) {
  int i, j;
  int prev = 0;
  for (i = 1; i <= n; i++) {
    while (prev + 1 < i && w[prev + 1] + KK <= w[i]) prev++;
    for (j = 1; j <= K; j++) {
      S[i][j] = S[i - 1][j];
    }
    S[i][1] = (S[i][1] + 1) % Mod;
    for (j = 2; j <= K; j++) {
      S[i][j] = (S[i][j] + S[prev][j - 1]) % Mod;
    }
  }
  return S[n][K];
}
int main() {
  int i;
  scanf("%d%d", &n, &K);
  for (i = 1; i <= n; i++) {
    scanf("%d", &w[i]);
  }
  sort(w + 1, w + n + 1);
  for (i = 0; i <= 100000 / (K - 1); i++) {
    Z[i] = Calc(i);
  }
  long long res = 0;
  for (i = 0; i <= 100000 / (K - 1); i++) {
    Z[i] = Z[i] - Z[i + 1];
    res = (res + 1ll * Z[i] * i) % Mod;
  }
  printf("%lld\n", res);
}
