#include <bits/stdc++.h>
using namespace std;
long long mulmod(long long, long long);
long long fastpower(long long, long long);
void getprime(vector<long long> &);
void getfact(vector<long long> &, vector<long long> &);
long long ncr(long long, long long, vector<long long> &, vector<long long> &);
void solve() {
  long long n, m, k, res = 0;
  cin >> n;
  while (n > 9) {
    res += (n / 10) * 10;
    n = (n / 10) + (n % 10);
  }
  res += n;
  cout << res << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
long long mulmod(long long a, long long b) {
  long long res = 0;
  a = a % 1000000007;
  while (b > 0) {
    if (b % 2 == 1) res = (res + a) % 1000000007;
    a = (a * 2) % 1000000007;
    b /= 2;
  }
  return res % 1000000007;
}
long long fastpower(long long a, long long n) {
  long long res = 1;
  while (n > 0) {
    if (n % 2 != 0) {
      res = (res * a) % 1000000007;
    }
    n /= 2;
    a = (a * a) % 1000000007;
  }
  return res;
}
void getprime(vector<bool> &prime) {
  prime[0] = false;
  prime[1] = false;
  for (long long i = 2; i <= 1000000; i++) {
    if (prime[i]) {
      for (long long j = i * i; j <= 1000000; j += i) prime[j] = false;
    }
  }
}
void getfact(vector<long long> &fact, vector<long long> &ifact) {
  fact[0] = 1;
  ifact[0] = 1;
  for (long long i = 1; i <= 1000000; i++) {
    fact[i] = (fact[i - 1] * i) % 1000000007;
    ifact[i] = (ifact[i - 1] * fastpower(i, 1000000007 - 2)) % 1000000007;
  }
}
long long ncr(long long n, long long r, vector<long long> &fact,
              vector<long long> &ifact) {
  long long res;
  res = (((fact[n] * ifact[r]) % 1000000007) * ifact[n - r]) % 1000000007;
  return res;
}

### Prompt

Your task is to create 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;
long long mulmod(long long, long long);
long long fastpower(long long, long long);
void getprime(vector<long long> &);
void getfact(vector<long long> &, vector<long long> &);
long long ncr(long long, long long, vector<long long> &, vector<long long> &);
void solve() {
  long long n, m, k, res = 0;
  cin >> n;
  while (n > 9) {
    res += (n / 10) * 10;
    n = (n / 10) + (n % 10);
  }
  res += n;
  cout << res << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
long long mulmod(long long a, long long b) {
  long long res = 0;
  a = a % 1000000007;
  while (b > 0) {
    if (b % 2 == 1) res = (res + a) % 1000000007;
    a = (a * 2) % 1000000007;
    b /= 2;
  }
  return res % 1000000007;
}
long long fastpower(long long a, long long n) {
  long long res = 1;
  while (n > 0) {
    if (n % 2 != 0) {
      res = (res * a) % 1000000007;
    }
    n /= 2;
    a = (a * a) % 1000000007;
  }
  return res;
}
void getprime(vector<bool> &prime) {
  prime[0] = false;
  prime[1] = false;
  for (long long i = 2; i <= 1000000; i++) {
    if (prime[i]) {
      for (long long j = i * i; j <= 1000000; j += i) prime[j] = false;
    }
  }
}
void getfact(vector<long long> &fact, vector<long long> &ifact) {
  fact[0] = 1;
  ifact[0] = 1;
  for (long long i = 1; i <= 1000000; i++) {
    fact[i] = (fact[i - 1] * i) % 1000000007;
    ifact[i] = (ifact[i - 1] * fastpower(i, 1000000007 - 2)) % 1000000007;
  }
}
long long ncr(long long n, long long r, vector<long long> &fact,
              vector<long long> &ifact) {
  long long res;
  res = (((fact[n] * ifact[r]) % 1000000007) * ifact[n - r]) % 1000000007;
  return res;
}

```