codeforces979A Pizza, Pizza, Pizza!!

最新推荐文章于 2024-10-30 02:03:01 发布

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最新推荐文章于 2024-10-30 02:03:01 发布

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Shiro在她的生日派对上需要将一个圆形比萨均匀地切成n+1份，以便她和她的朋友们都能享用到相同大小和形状的比萨片。本篇介绍了如何根据参与派对的朋友数量确定最少的切割次数。

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Katie, Kuro and Shiro are best friends. They have known each other since kindergarten. That's why they often share everything with each other and work together on some very hard problems.

Today is Shiro's birthday. She really loves pizza so she wants to invite her friends to the pizza restaurant near her house to celebrate her birthday, including her best friends Katie and Kuro.

She has ordered a very big round pizza, in order to serve her many friends. Exactly $n$

of Shiro's friends are here. That's why she has to divide the pizza into

n + 1

slices (Shiro also needs to eat). She wants the slices to be exactly the same size and shape. If not, some of her friends will get mad and go home early, and the party will be over.

Shiro is now hungry. She wants to cut the pizza with minimum of straight cuts. A cut is a straight segment, it might have ends inside or outside the pizza. But she is too lazy to pick up the calculator.

As usual, she will ask Katie and Kuro for help. But they haven't come yet. Could you help Shiro with this problem?

Input

A single line contains one non-negative integer $n$

(

0 \leq n \leq 10^{18}

) — the number of Shiro's friends. The circular pizza has to be sliced into

n + 1

pieces.

Output

A single integer — the number of straight cuts Shiro needs.

Examples

Input

Copy

Output

Copy

Input

Copy

Output

Copy

Note

To cut the round pizza into quarters one has to make two cuts through the center with angle $90^{\circ}$

between them.

To cut the round pizza into five equal parts one has to make five cuts.

情况一：没有朋友，需要1块，切0刀。

情况二：朋友为奇数，需要偶数n块，切n/2刀。

情况三：朋友为偶数，需要奇数n块，切n刀。

注意溢出，使用long long。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <queue>

using namespace std;


int main(){

        long long n;
        cin>>n;
        n=n+1;
        long long ans;
        if(n==1){
            ans=0;
            cout<<ans<<endl;
            return 0;
        }
        if(n%2==0){
            ans=n/2;
        }else{
            ans=n;
        }

        cout<<ans<<endl;


    return 0;
}