#include <bits/stdc++.h>
using namespace std;
void func(void) {
  freopen("input.c", "r", stdin);
  freopen("output.c", "w", stdout);
}
void print(vector<long long> &v) {
  cout << v.size() << endl;
  for (int i = 0; i < v.size(); i++) {
    printf("%lld ", v[i]);
  }
  printf("\n");
}
void print(vector<pair<long long, long long> > &v) {
  cout << v.size() << endl;
  for (int i = 0; i < v.size(); i++) {
    printf("%lld %lld\n", v[i].first, v[i].second);
  }
}
void print(double d) { cout << fixed << setprecision(10) << d << endl; }
void print(string s, double d) {
  cout << s << " ";
  cout << fixed << setprecision(10) << d << endl;
}
int main() {
  long long n, q, i, j = 0, temp, t, k, ans = 0, sum = 0, x, y, z, cnt = 0, m,
                     fg = 0, mx = 0, mx1 = 0, mn = 8000000000000000000,
                     mn1 = 8000000000000000000;
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld", &x);
    ans = 0;
    while (x >= 10) {
      ans += (x / 10) * 10;
      x = (x / 10) + (x % 10);
    }
    ans += x;
    cout << ans << endl;
  }
}

### Prompt

In cpp, your task is to solve 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;
void func(void) {
  freopen("input.c", "r", stdin);
  freopen("output.c", "w", stdout);
}
void print(vector<long long> &v) {
  cout << v.size() << endl;
  for (int i = 0; i < v.size(); i++) {
    printf("%lld ", v[i]);
  }
  printf("\n");
}
void print(vector<pair<long long, long long> > &v) {
  cout << v.size() << endl;
  for (int i = 0; i < v.size(); i++) {
    printf("%lld %lld\n", v[i].first, v[i].second);
  }
}
void print(double d) { cout << fixed << setprecision(10) << d << endl; }
void print(string s, double d) {
  cout << s << " ";
  cout << fixed << setprecision(10) << d << endl;
}
int main() {
  long long n, q, i, j = 0, temp, t, k, ans = 0, sum = 0, x, y, z, cnt = 0, m,
                     fg = 0, mx = 0, mx1 = 0, mn = 8000000000000000000,
                     mn1 = 8000000000000000000;
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld", &x);
    ans = 0;
    while (x >= 10) {
      ans += (x / 10) * 10;
      x = (x / 10) + (x % 10);
    }
    ans += x;
    cout << ans << endl;
  }
}

```