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;
int k, n;
const int maxn = 1e3 + 10;
const int mod = 998244353;
int a[maxn];
int dp[maxn][maxn];
int solve(int limit) {
  for (int j = 2; j <= k; ++j) {
    int now_v = 0, now_i = j - 2;
    for (int i = j; i <= n; ++i) {
      while (a[now_i + 1] + limit <= a[i] && now_i + 1 < i) {
        now_i++;
        now_v = (now_v + dp[now_i][j - 1]) % mod;
      }
      dp[i][j] = now_v;
    }
  }
  int ans = 0;
  for (int i = k; i <= n; ++i) ans = (ans + dp[i][k]) % mod;
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  sort(a + 1, a + 1 + n);
  int ans = 0;
  for (int i = 1; i <= n; ++i) dp[i][1] = 1;
  for (int i = 1; i <= (a[n] - a[1]) / (k - 1); ++i)
    ans = (ans + solve(i)) % mod;
  cout << ans << endl;
  return 0;
}
