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 mod = 998244353;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  int a[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  int upper_x = a[n - 1] / (k - 1);
  long long ans = 0;
  for (int x = 1; x <= upper_x; x++) {
    long long dp[n][k + 1];
    long long psum[n][k + 1];
    int last[n];
    for (int i = 0; i < n; i++) {
      last[i] = 0;
      dp[i][1] = 1;
      psum[i][1] = i + 1;
      for (int j = 2; j <= k; j++) {
        dp[i][j] = 0;
        psum[i][j] = 0;
      }
    }
    for (int i = 1; i < n; i++) {
      if (a[i] >= a[i - 1] + x)
        last[i] = i - 1;
      else {
        last[i] = max(last[i - 1], 0);
        while (a[last[i]] + x <= a[i]) last[i]++;
        if (last[i] >= 0) last[i]--;
      }
    }
    last[0] = -1;
    for (int i = 0; i < n; i++) {
      for (int j = 2; j <= k; j++) {
        if (last[i] == -1)
          dp[i][j] = 0;
        else {
          dp[i][j] = psum[last[i]][j - 1];
        }
        psum[i][j] = dp[i][j];
        if (i) psum[i][j] += psum[i - 1][j];
        dp[i][j] %= mod;
        psum[i][j] %= mod;
      }
      ans += dp[i][k];
      ans %= mod;
    }
  }
  cout << ans << '\n';
  return 0;
}
