#include <bits/stdc++.h>
using namespace std;
const long long INF = INT_MAX;
const long long LINF = LLONG_MAX;
const long long MOD = 1000000000 + 7;
const double EPS = 1e-9;
const double PI = acos(-1);
const int di[]{0, -1, 0, 1, -1, -1, 1, 1};
const int dj[]{1, 0, -1, 0, 1, -1, -1, 1};
long long gcd(long long a, long long b) {
  long long r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
long long power(long long x, long long n, long long mod = MOD) {
  long long ret = 1;
  while (n > 0) {
    if (n & 1) {
      (ret *= x) %= mod;
    }
    (x *= x) %= mod;
    n >>= 1;
  }
  return ret;
}
vector<long long> make_power(int n, long long base) {
  long long num = 1;
  vector<long long> ret;
  for (int i = 0; i <= n; ++i) {
    ret.push_back(num);
    num *= base;
  }
  return ret;
}
long long digit_sum(long long n) {
  long long dig;
  long long sum = 0;
  while (n) {
    dig = n % 10;
    sum = sum + dig;
    n = n / 10;
  }
  return sum;
}
long long s;
void input() {
  cin >> s;
  return;
}
void solve() {
  long long ans = 0;
  while (s >= 10) {
    long long d = s / 10;
    ans += d * 10;
    s -= 10 * d;
    s += d;
  }
  ans += s;
  cout << ans << endl;
  return;
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  int t;
  cin >> t;
  for (int i = (0); i < (t); i++) {
    input();
    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;
const long long INF = INT_MAX;
const long long LINF = LLONG_MAX;
const long long MOD = 1000000000 + 7;
const double EPS = 1e-9;
const double PI = acos(-1);
const int di[]{0, -1, 0, 1, -1, -1, 1, 1};
const int dj[]{1, 0, -1, 0, 1, -1, -1, 1};
long long gcd(long long a, long long b) {
  long long r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
long long power(long long x, long long n, long long mod = MOD) {
  long long ret = 1;
  while (n > 0) {
    if (n & 1) {
      (ret *= x) %= mod;
    }
    (x *= x) %= mod;
    n >>= 1;
  }
  return ret;
}
vector<long long> make_power(int n, long long base) {
  long long num = 1;
  vector<long long> ret;
  for (int i = 0; i <= n; ++i) {
    ret.push_back(num);
    num *= base;
  }
  return ret;
}
long long digit_sum(long long n) {
  long long dig;
  long long sum = 0;
  while (n) {
    dig = n % 10;
    sum = sum + dig;
    n = n / 10;
  }
  return sum;
}
long long s;
void input() {
  cin >> s;
  return;
}
void solve() {
  long long ans = 0;
  while (s >= 10) {
    long long d = s / 10;
    ans += d * 10;
    s -= 10 * d;
    s += d;
  }
  ans += s;
  cout << ans << endl;
  return;
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  int t;
  cin >> t;
  for (int i = (0); i < (t); i++) {
    input();
    solve();
  }
  return 0;
}

```