Codeforces Round #451 (Div. 2) A. Rounding【分类讨论/易错】

A. Rounding

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Input

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

Output

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

Examples

input

5

output

0

input

113

output

110

input

1000000000

output

1000000000

input

5432359

output

5432360

Note

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.

【分析】：注释

【代码】：

#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 ;
}

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