#include <bits/stdc++.h>
using namespace std;
long long GCD(long long, long long);
long long LCM(long long, long long);
long long power(int, int);
template <typename T>
inline T mmod(T a, T b, T mod) {
  return ((a % mod) * (b % mod)) % mod;
}
template <typename T>
inline T smod(T a, T b, T mod) {
  return ((a % mod) + (b % mod)) % mod;
}
template <typename T>
inline void MIN(T& x, T y) {
  if (x > y) x = y;
}
template <typename T>
inline void MAX(T& x, T y) {
  if (x < y) x = y;
}
long long nCrModp(long long n, long long r, long long p) {
  long long C[r + 1];
  memset(C, 0, sizeof(C));
  C[0] = 1;
  for (long long i = 1; i <= n; i++) {
    for (long long j = (i < r ? i : r); j > 0; j--)
      C[j] = (C[j] + C[j - 1]) % p;
  }
  return C[r];
}
long long d, x, y;
void extendedEuclid(long long A, long long B) {
  if (B == 0) {
    d = A;
    x = 1;
    y = 0;
  } else {
    extendedEuclid(B, A % B);
    int temp = x;
    x = y;
    y = temp - (A / B) * y;
  }
}
long long GCD(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return GCD(b, a % b);
}
long long LCM(long long a, long long b) {
  return ((a < b ? b : a) / GCD(a, b)) * (a < b ? a : b);
}
long long fpower(long long a, long long n) {
  unsigned long long int result = 1, x = a;
  while (n > 0) {
    if (n % 2 == 1) result = result * x;
    x = x * x;
    n = n / 2;
  }
  return result;
}
long long fmpower(long long int base, long long int exp, long long md) {
  long long res = 1;
  while (exp > 0) {
    if (exp % 2 == 1) res = (res * base) % md;
    base = (base * base) % md;
    exp /= 2;
  }
  return res;
}
signed main() {
  int t;
  cin >> t;
  while (t--) {
    long long int n, ans = 0, m = 10LL;
    cin >> n;
    ans = m * (n / 9) + n % 9 - (bool)(n % 9 == 0);
    cout << ans << "\n";
  }
}

### Prompt

Develop a solution in cpp to the problem described below:

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 GCD(long long, long long);
long long LCM(long long, long long);
long long power(int, int);
template <typename T>
inline T mmod(T a, T b, T mod) {
  return ((a % mod) * (b % mod)) % mod;
}
template <typename T>
inline T smod(T a, T b, T mod) {
  return ((a % mod) + (b % mod)) % mod;
}
template <typename T>
inline void MIN(T& x, T y) {
  if (x > y) x = y;
}
template <typename T>
inline void MAX(T& x, T y) {
  if (x < y) x = y;
}
long long nCrModp(long long n, long long r, long long p) {
  long long C[r + 1];
  memset(C, 0, sizeof(C));
  C[0] = 1;
  for (long long i = 1; i <= n; i++) {
    for (long long j = (i < r ? i : r); j > 0; j--)
      C[j] = (C[j] + C[j - 1]) % p;
  }
  return C[r];
}
long long d, x, y;
void extendedEuclid(long long A, long long B) {
  if (B == 0) {
    d = A;
    x = 1;
    y = 0;
  } else {
    extendedEuclid(B, A % B);
    int temp = x;
    x = y;
    y = temp - (A / B) * y;
  }
}
long long GCD(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return GCD(b, a % b);
}
long long LCM(long long a, long long b) {
  return ((a < b ? b : a) / GCD(a, b)) * (a < b ? a : b);
}
long long fpower(long long a, long long n) {
  unsigned long long int result = 1, x = a;
  while (n > 0) {
    if (n % 2 == 1) result = result * x;
    x = x * x;
    n = n / 2;
  }
  return result;
}
long long fmpower(long long int base, long long int exp, long long md) {
  long long res = 1;
  while (exp > 0) {
    if (exp % 2 == 1) res = (res * base) % md;
    base = (base * base) % md;
    exp /= 2;
  }
  return res;
}
signed main() {
  int t;
  cin >> t;
  while (t--) {
    long long int n, ans = 0, m = 10LL;
    cin >> n;
    ans = m * (n / 9) + n % 9 - (bool)(n % 9 == 0);
    cout << ans << "\n";
  }
}

```