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 A[1005];
long long res, mod = 998244353ll;
long long dp[1005][1005], pre;
int main() {
  scanf("%d %d", &n, &k);
  for (int i = (1); i <= (n); i++) scanf("%d", &A[i]);
  sort(A + 1, A + 1 + n);
  int atmost = (A[n] - A[1]) / (k - 1);
  for (int req = (0); req <= (atmost); req++) {
    for (int i = (1); i <= (n); i++) dp[1][i] = i;
    for (int i = (2); i <= (k); i++) {
      int x = 0;
      for (int j = (1); j <= (n); j++) {
        while (A[j] - A[x + 1] >= req && x + 1 < j) x++;
        dp[i][j] = (dp[i][j - 1] + dp[i - 1][x]) % mod;
      }
    }
    long long cur = dp[k][n];
    if (req > 0) {
      res = res + (req - 1) * ((pre - cur + mod) % mod);
      res %= mod;
    }
    pre = cur;
    for (int i = (1); i <= (k); i++)
      for (int j = (1); j <= (n); j++) dp[i][j] = 0;
  }
  res = res + atmost * pre;
  res %= mod;
  printf("%lld\n", res);
  return 0;
}
