#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int t;
  cin >> t;
  vector<long long int> ans;
  for (long long int j = 0; j < t; j++) {
    long long s, r;
    cin >> s;
    r = s;
    if (s < 10)
      ans.push_back(s);
    else {
      long long sum = 0;
      while (r >= 10) {
        long long temp;
        temp = r / 10;
        sum = sum + temp * 10;
        r = temp + r % 10;
      }
      sum = sum + r;
      ans.push_back(sum);
    }
  }
  for (long long int j = 0; j < t; j++) {
    cout << ans[j] << endl;
  }
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

Mishka wants to buy some food in the nearby shop. Initially, he has s burles on his card. 

Mishka can perform the following operation any number of times (possibly, zero): choose some positive integer number 1 ≤ x ≤ s, buy food that costs exactly x burles and obtain ⌊x/10⌋ burles as a cashback (in other words, Mishka spends x burles and obtains ⌊x/10⌋ back). The operation ⌊a/b⌋ means a divided by b rounded down.

It is guaranteed that you can always buy some food that costs x for any possible value of x.

Your task is to say the maximum number of burles Mishka can spend if he buys food optimally.

For example, if Mishka has s=19 burles then the maximum number of burles he can spend is 21. Firstly, he can spend x=10 burles, obtain 1 burle as a cashback. Now he has s=10 burles, so can spend x=10 burles, obtain 1 burle as a cashback and spend it too.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases.

The next t lines describe test cases. Each test case is given on a separate line and consists of one integer s (1 ≤ s ≤ 10^9) — the number of burles Mishka initially has.

Output

For each test case print the answer on it — the maximum number of burles Mishka can spend if he buys food optimally.

Example

Input


6
1
10
19
9876
12345
1000000000


Output


1
11
21
10973
13716
1111111111

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int t;
  cin >> t;
  vector<long long int> ans;
  for (long long int j = 0; j < t; j++) {
    long long s, r;
    cin >> s;
    r = s;
    if (s < 10)
      ans.push_back(s);
    else {
      long long sum = 0;
      while (r >= 10) {
        long long temp;
        temp = r / 10;
        sum = sum + temp * 10;
        r = temp + r % 10;
      }
      sum = sum + r;
      ans.push_back(sum);
    }
  }
  for (long long int j = 0; j < t; j++) {
    cout << ans[j] << endl;
  }
}

```