#include <bits/stdc++.h>
int main() {
  long long int N, T, U, i, j, x, y, k, s, m, n;
  scanf("%lld", &N);
  for (i = 1; i <= N; i++) {
    scanf("%lld", &T);
    s = 0;
    x = T;
    while (x > 0) {
      s = s + T - (T % 10);
      T = (T / 10) + (T % 10);
      x = T / 10;
    }
    printf("%lld\n", T + s);
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
int main() {
  long long int N, T, U, i, j, x, y, k, s, m, n;
  scanf("%lld", &N);
  for (i = 1; i <= N; i++) {
    scanf("%lld", &T);
    s = 0;
    x = T;
    while (x > 0) {
      s = s + T - (T % 10);
      T = (T / 10) + (T % 10);
      x = T / 10;
    }
    printf("%lld\n", T + s);
  }
  return 0;
}

```