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;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
const int N = 1000 + 7;
const int M = 998244353;
int dp[N][N];
int pref[N][N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  int ans = 0;
  for (int x = 1; x <= 100000 / (k - 1); x++) {
    for (int i = 0; i < n; i++) {
      for (int j = 0; j <= k; j++) {
        dp[i][j] = 0;
      }
    }
    int y = 0;
    for (int i = 0; i < n; i++) {
      while (y >= 0 && a[i] - a[y] < x) y--;
      y = min(y, i - 1);
      while (y + 1 < i && a[i] - a[y + 1] >= x) y++;
      dp[i][1] = 1;
      for (int j = 1; j < k; j++) {
        if (y >= 0) dp[i][j + 1] = pref[y][j];
      }
      for (int j = 1; j < k; j++) {
        pref[i][j] = (i ? pref[i - 1][j] : 0) + dp[i][j];
        if (pref[i][j] >= M) pref[i][j] -= M;
      }
      ans += dp[i][k];
      if (ans >= M) ans -= M;
    }
  }
  cout << ans << '\n';
}
