Vasya has a non-negative integer n. He wants to round it to nearest integer, which ends up with 0. If n already ends up with 0, Vasya considers it already rounded.

For example, if n = 4722 answer is 4720. If n = 5 Vasya can round it to 0 or to 10. Both ways are correct.

For given n find out to which integer will Vasya round it.

The first line contains single integer n (0 ≤ n ≤ 109) — number that Vasya has.

Print result of rounding n. Pay attention that in some cases answer isn't unique. In that case print any correct answer.

In the first example n = 5. Nearest integers, that ends up with zero are 0 and 10. Any of these answers is correct, so you can print 0 or10.

#include<cstdio>

#include<iostream>

#include<cmath>

using namespace std;

const int m=;

int n,x,l;

long long ans;

int main()

{

    while(~scanf("%d",&n))

    {

        int t=n%;

        if(t==)//尾数为0

        {

            return *printf("%d\n",n);

        }

        else if(n<=)//不大于4的个位数

        {

            return *printf("0\n");

        }

        else if(n>=&&n<=)//大于4的个位数

        {

            return *printf("10\n");

        }

        else if(t>=)//尾数大于4

        {

            return *printf("%d\n",n+(-t));

        }

        else if(t<=)//尾数小于4

        {

            return *printf("%d\n",n-t);

        }

    }

    return ;

}