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;
const int N = 1005, Mx = 1e5 + 10, mod = 998244353;
int n, k, ans, a[N], c[N * N], dp[N][N];
void solve(int x) {
  dp[0][0] = 1;
  for (int i = 1; i < n + 1; i++)
    for (int j = 0; j < k + 1; j++) {
      dp[i][j] = dp[i - 1][j];
      if (j) dp[i][j] = (dp[i][j] + dp[c[max(a[i] - x, 0)]][j - 1]) % mod;
      ;
    }
}
int main() {
  cin >> n >> k;
  for (int i = 1; i < n + 1; i++) {
    scanf("%d", &a[i]);
    ;
    a[i]++;
  }
  sort(a + 1, a + 1 + n);
  for (int i = 1; i < n + 1; i++) c[a[i]] = i;
  for (int i = 1; i < Mx; i++) c[i] = max(c[i], c[i - 1]);
  for (int i = 1; i < Mx / (k - 1) + 1; i++) {
    solve(i);
    ans = (ans + dp[n][k]) % mod;
    ;
  }
  cout << ans;
}
