D. Vus the Cossack and Numbers

D. Vus the Cossack and Numbers

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vus the Cossack has nn real numbers aiai. It is known that the sum of all numbers is equal to 00. He wants to choose a sequence bb the size of which is nn such that the sum of all numbers is 00 and each bibi is either ⌊ai⌋⌊ai⌋ or ⌈ai⌉⌈ai⌉. In other words, bibi equals aiai rounded up or down. It is not necessary to round to the nearest integer.

For example, if a=[4.58413,1.22491,−2.10517,−3.70387]a=[4.58413,1.22491,−2.10517,−3.70387], then bb can be equal, for example, to [4,2,−2,−4][4,2,−2,−4].

Note that if aiai is an integer, then there is no difference between ⌊ai⌋⌊ai⌋ and ⌈ai⌉⌈ai⌉, bibi will always be equal to aiai.

Help Vus the Cossack find such sequence!

Input

The first line contains one integer nn (1≤n≤1051≤n≤105) — the number of numbers.

Each of the next nn lines contains one real number aiai (|ai|<105|ai|<105). It is guaranteed that each aiai has exactly 55 digits after the decimal point. It is guaranteed that the sum of all the numbers is equal to 00.

Output

In each of the next nn lines, print one integer bibi. For each ii, |ai−bi|<1|ai−bi|<1 must be met.

If there are multiple answers, print any.

Examples

input

Copy

4
4.58413
1.22491
-2.10517
-3.70387

output

Copy

4
2
-2
-4

input

Copy

5
-6.32509
3.30066
-0.93878
2.00000
1.96321

output

Copy

-6
3
-1
2
2

Note

The first example is explained in the legend.

In the second example, we can round the first and fifth numbers up, and the second and third numbers down. We can round the fourth number neither up, nor down.

=========================================================================

贪心一下，先都向下取整，求出来和，小于零就加1，加1的情况只在a[i]不等于b[i]的时候

# include
# include
# include
 
using namespace std;
 
 
int main()
{
 
 
  int n;
  cin>>n;
  double a[100000+10];
 
  double b[100000+10];
  double sum=0;
  for(int i=1;i<=n;i++)
  {
      cin>>a[i];
 
      b[i]=floor(a[i]);
 
 
      sum+=b[i];
  }
 
 
  for(int i=1;i<=n;i++)
  {
      if(abs(a[i]-b[i])<1e-6)
        continue;
 
 
      if(sum<0)
      {
          sum++;
 
          b[i]++;
      }
      else
      {
          break;
      }
  }
 
  for(int i=1;i<=n;i++)
  {
      cout<
  }
 
 
 
    return 0;
}