#include <bits/stdc++.h>
using namespace std;
const long long N = 998244353;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long binpow(long long a, int n) {
  if (!n) return 1;
  if (n % 2) return binpow(a, n - 1) * a % N;
  long long bp = binpow(a, n / 2);
  return bp * bp % N;
}
void solve() {
  int n;
  cin >> n;
  vector<long long> c(n), d(n + 1);
  for (int i = 0; i < n; ++i) cin >> c[i];
  sort(c.begin(), c.end());
  for (int i = 1; i <= n; ++i) {
    d[i] = (d[i - 1] + c[i - 1]) % N;
  }
  vector<long long> ans(n);
  for (int i = 0; i < n - 1; ++i) {
    for (int j = n - i - 1; j > -1; j -= i + 1) {
      ans[i] += d[j];
    }
    ans[i] %= N;
  }
  long long rev = binpow(n, N - 2);
  for (int i = 0; i < n; ++i) {
    cout << ans[i] * rev % N << " ";
  }
  cout << endl;
}
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

You are creating a level for a video game. The level consists of n rooms placed in a circle. The rooms are numbered 1 through n. Each room contains exactly one exit: completing the j-th room allows you to go the (j+1)-th room (and completing the n-th room allows you to go the 1-st room).

You are given the description of the multiset of n chests: the i-th chest has treasure value c_i.

Each chest can be of one of two types: 

  * regular chest — when a player enters a room with this chest, he grabs the treasure and proceeds to the next room; 
  * mimic chest — when a player enters a room with this chest, the chest eats him alive, and he loses. 



The player starts in a random room with each room having an equal probability of being chosen. The players earnings is equal to the total value of treasure chests he'd collected before he lost.

You are allowed to choose the order the chests go into the rooms. For each k from 1 to n place the chests into the rooms in such a way that:

  * each room contains exactly one chest; 
  * exactly k chests are mimics; 
  * the expected value of players earnings is minimum possible. 



Please note that for each k the placement is chosen independently.

It can be shown that it is in the form of P/Q where P and Q are non-negative integers and Q ≠ 0. Report the values of P ⋅ Q^{-1} \pmod {998244353}.

Input

The first contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the number of rooms and the number of chests.

The second line contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ 10^6) — the treasure values of each chest.

Output

Print n integers — the k -th value should be equal to the minimum possible expected value of players earnings if the chests are placed into the rooms in some order and exactly k of the chests are mimics.

It can be shown that it is in the form of P/Q where P and Q are non-negative integers and Q ≠ 0. Report the values of P ⋅ Q^{-1} \pmod {998244353}.

Examples

Input


2
1 2


Output


499122177 0 


Input


8
10 4 3 6 5 10 7 5


Output


499122193 249561095 249561092 873463811 499122178 124780545 623902721 0 

Note

In the first example the exact values of minimum expected values are: \frac 1 2, \frac 0 2.

In the second example the exact values of minimum expected values are: \frac{132} 8, \frac{54} 8, \frac{30} 8, \frac{17} 8, \frac{12} 8, \frac 7 8, \frac 3 8, \frac 0 8.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long N = 998244353;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long binpow(long long a, int n) {
  if (!n) return 1;
  if (n % 2) return binpow(a, n - 1) * a % N;
  long long bp = binpow(a, n / 2);
  return bp * bp % N;
}
void solve() {
  int n;
  cin >> n;
  vector<long long> c(n), d(n + 1);
  for (int i = 0; i < n; ++i) cin >> c[i];
  sort(c.begin(), c.end());
  for (int i = 1; i <= n; ++i) {
    d[i] = (d[i - 1] + c[i - 1]) % N;
  }
  vector<long long> ans(n);
  for (int i = 0; i < n - 1; ++i) {
    for (int j = n - i - 1; j > -1; j -= i + 1) {
      ans[i] += d[j];
    }
    ans[i] %= N;
  }
  long long rev = binpow(n, N - 2);
  for (int i = 0; i < n; ++i) {
    cout << ans[i] * rev % N << " ";
  }
  cout << endl;
}
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```