#include <bits/stdc++.h>
using namespace std;
template <class T>
ostream& operator<<(ostream& os, vector<T> V) {
  os << "[ ";
  for (auto v : V) os << v << " ";
  return os << " ]";
}
template <class T>
ostream& operator<<(ostream& os, set<T> V) {
  os << "[ ";
  for (auto v : V) os << v << " ";
  return os << " ]";
}
template <class T, class R>
ostream& operator<<(ostream& os, map<T, R> V) {
  os << "[ ";
  for (auto v : V) os << "(" << v.first << ":" << v.second << ")";
  return os << " ]";
}
template <class T, class R>
ostream& operator<<(ostream& os, pair<T, R> V) {
  return os << "(" << V.first << "," << V.second << ")";
}
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
clock_t clk = clock();
const long long int Mod = 1e9 + 7;
const long long int MAXN = 1e5;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(25);
  long long int t;
  cin >> t;
  while (t--) {
    long long int s;
    cin >> s;
    long long int ans = 0;
    while (s > 9) {
      ans += (s / 10) * 10;
      s = s / 10 + s % 10;
    }
    ans += s;
    cout << ans << "\n";
  }
  return 0;
}

### Prompt

Please formulate 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;
template <class T>
ostream& operator<<(ostream& os, vector<T> V) {
  os << "[ ";
  for (auto v : V) os << v << " ";
  return os << " ]";
}
template <class T>
ostream& operator<<(ostream& os, set<T> V) {
  os << "[ ";
  for (auto v : V) os << v << " ";
  return os << " ]";
}
template <class T, class R>
ostream& operator<<(ostream& os, map<T, R> V) {
  os << "[ ";
  for (auto v : V) os << "(" << v.first << ":" << v.second << ")";
  return os << " ]";
}
template <class T, class R>
ostream& operator<<(ostream& os, pair<T, R> V) {
  return os << "(" << V.first << "," << V.second << ")";
}
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
clock_t clk = clock();
const long long int Mod = 1e9 + 7;
const long long int MAXN = 1e5;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(25);
  long long int t;
  cin >> t;
  while (t--) {
    long long int s;
    cin >> s;
    long long int ans = 0;
    while (s > 9) {
      ans += (s / 10) * 10;
      s = s / 10 + s % 10;
    }
    ans += s;
    cout << ans << "\n";
  }
  return 0;
}

```