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>
const int MOD = 998244353;
int sum(int a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
  return a;
}
const int N = 1001;
const int MAXA = 1e5;
int a[N];
int dp[N][N];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; ++i) scanf("%d", a + i);
  std::sort(a, a + n);
  int ans = 0;
  for (int x = 1; x * (k - 1) <= MAXA; ++x) {
    for (int i = 0; i <= n; ++i) dp[i][0] = 1;
    for (int t = 1; t <= k; ++t) dp[0][t] = 0;
    int pos = 0;
    for (int i = 0; i < n; ++i) {
      for (int t = 1; t <= k; ++t) {
        while (a[pos] + x <= a[i]) ++pos;
        dp[i + 1][t] = sum(dp[i][t], dp[pos][t - 1]);
      }
    }
    ans = sum(ans, dp[n][k]);
  }
  printf("%d", ans);
}
