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 maxn = 1000 + 10;
const int p = 998244353;
int dp[maxn][maxn];
int n, k;
int a[maxn];
int s[maxn][maxn];
int ans;
int solve(int x) {
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= n; j++) {
      dp[i][j] = 0;
      s[i][j] = 0;
    }
  }
  for (int j = 1; j <= n; j++) dp[1][j] = 1;
  for (int i = 1; i <= k; i++) {
    int l = 1;
    for (int j = 1; j <= n; j++) {
      while (a[j] - a[l] >= x) l++;
      dp[i][j] = (dp[i][j] + s[i - 1][l - 1]) % p;
      s[i][j] = (s[i][j - 1] + dp[i][j]) % p;
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) ans = (ans + dp[k][i]) % p;
  return ans;
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= (a[n]) / (k - 1); i++) {
    ans = (ans + solve(i)) % p;
  }
  cout << ans << endl;
  return 0;
}
