#include <bits/stdc++.h>
using namespace std;
bool is_vowel(char x) {
  if (x == 'a' or x == 'e' or x == 'i' or x == 'o' or x == 'u') return true;
  if (x == 'A' or x == 'E' or x == 'I' or x == 'O' or x == 'U') return true;
  if (x == 'y' or x == 'Y') return true;
  return false;
}
bool is_palindrome(string s) {
  int len = s.length();
  int n = len - 1;
  for (int i = 0; i < len; ++i, --n) {
    if (s[i] != s[n]) return false;
  }
  return true;
}
string to_str(long long x) {
  string ans = "";
  while (x != 0) {
    int d = x % 10;
    char c = (int)d + 48;
    ans.push_back(c);
    x = x / 10;
  }
  reverse(ans.begin(), ans.end());
  return ans;
}
bool is_prime(long long x) {
  if (x == 1 or x == 0) return false;
  if (x == 2) return true;
  for (long long i = 2; i * i <= x; ++i) {
    if (x % i == 0) return false;
  }
  return true;
}
int car_todig(char x) { return x - 48; }
char dig_tocar(int x) { return x + 48; }
bool is_upper(char x) {
  if (x >= 'A' and x <= 'Z')
    return true;
  else
    return false;
}
bool is_lower(char x) {
  if (x >= 'a' and x <= 'z')
    return true;
  else
    return false;
}
bool is_digit(char x) {
  if (x >= '0' and x <= '9')
    return true;
  else
    return false;
}
void solve() {
  long long n;
  cin >> n;
  long long ans = n;
  long long cnt = 0;
  while (n / 10 > 0) {
    long long x = n / 10;
    cnt += x;
    n = n - (x * 10);
    n += x;
  }
  ans += cnt;
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t = 1;
  cin >> t;
  for (int i = 1; i <= t; ++i) {
    solve();
  }
  return 0;
}

### Prompt

Generate 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;
bool is_vowel(char x) {
  if (x == 'a' or x == 'e' or x == 'i' or x == 'o' or x == 'u') return true;
  if (x == 'A' or x == 'E' or x == 'I' or x == 'O' or x == 'U') return true;
  if (x == 'y' or x == 'Y') return true;
  return false;
}
bool is_palindrome(string s) {
  int len = s.length();
  int n = len - 1;
  for (int i = 0; i < len; ++i, --n) {
    if (s[i] != s[n]) return false;
  }
  return true;
}
string to_str(long long x) {
  string ans = "";
  while (x != 0) {
    int d = x % 10;
    char c = (int)d + 48;
    ans.push_back(c);
    x = x / 10;
  }
  reverse(ans.begin(), ans.end());
  return ans;
}
bool is_prime(long long x) {
  if (x == 1 or x == 0) return false;
  if (x == 2) return true;
  for (long long i = 2; i * i <= x; ++i) {
    if (x % i == 0) return false;
  }
  return true;
}
int car_todig(char x) { return x - 48; }
char dig_tocar(int x) { return x + 48; }
bool is_upper(char x) {
  if (x >= 'A' and x <= 'Z')
    return true;
  else
    return false;
}
bool is_lower(char x) {
  if (x >= 'a' and x <= 'z')
    return true;
  else
    return false;
}
bool is_digit(char x) {
  if (x >= '0' and x <= '9')
    return true;
  else
    return false;
}
void solve() {
  long long n;
  cin >> n;
  long long ans = n;
  long long cnt = 0;
  while (n / 10 > 0) {
    long long x = n / 10;
    cnt += x;
    n = n - (x * 10);
    n += x;
  }
  ans += cnt;
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t = 1;
  cin >> t;
  for (int i = 1; i <= t; ++i) {
    solve();
  }
  return 0;
}

```