Codeforces Round #571 (Div. 2)-D. Vus the Cossack and Numbers

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.

题意:原来的浮点型数组之和为0，现在让你进行向上或者向下取整，使得变化后的数组之和也是0

思路：把浮点型数组进行一种取整方式，然后把小数部分保留一下，看最后的小数部分的进行四舍五入后的结果，然后根据最后小数部分的和进行不断地进行改变取整方式就可以了

代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<vector>
#include<cmath>
 
const int maxn=1e5+;
typedef long long ll;
using namespace std;
double a[maxn];
int b[maxn];
 
int  main()
{
    int n;
    cin>>n;
    double s=;
    for(int t=;t<n;t++)
    {
        scanf("%lf",&a[t]);
        b[t]=a[t];
        s+=(a[t]-b[t]);
    }
    int ss=round(s);
    if(ss>)
    {
 
    for(int t=;t<n;t++)
    {
        if(ss==)
        {
            break;
        }
        if(a[t]>b[t])
        {
            b[t]++;
            ss--;
        }
    }
    }
    if(ss<)
    {
    for(int t=;t<n;t++)
    {
        if(ss==)
        {
            break;
        }
        if(a[t]<b[t])
        {
            b[t]--;
            ss++;
        }
    }
    }
    for(int t=;t<n;t++)
    {
        cout<<b[t]<<endl;
    }
 
    return ;
}