Problem:
Score : 100 points

Problem Statement

You are given a positive integer N.
Find the minimum positive integer divisible by both 2 and N.

Constraints

1 \leq N \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N

Output

Print the minimum positive integer divisible by both 2 and N.

Sample Input 1

3

Sample Output 1

6

6 is divisible by both 2 and 3.
Also, there is no positive integer less than 6 that is divisible by both 2 and 3.
Thus, the answer is 6.

Sample Input 2

10

Sample Output 2

10

Sample Input 3

999999999

Sample Output 3

1999999998

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include "bits/stdc++.h"
using namespace std;

/*第二引数で第一引数を割ったときの切り上げの計算*/
long long int maxtime(long long int x, long long int y) {
	return(x + y - 1) / y;

}
/*最大公約数*/
long long int lcm(long long int number1, long long int number2) {
	long long int m = number1;
	long long int n = number2;

	if (number2 > number1) {
		m = number2;
		n = number1;
	}
	long long int s = -1;
	while (s != 0) {
		s = m % n;
		m = n;
		n = s;
	}
	return m;
}
/*最大公倍数*/
long long int gcd(long long int number1, long long int number2) {
	long long int m = number1;
	long long int n = number2;
	return m / lcm(m, n) * n;
}
int saisyo[100001] = {};
int nibanme[100001] = {};
int main() {
	int N;
	cin >> N;
	if (N % 2 == 0) {
		cout << N;
	}
	else {
		cout << N * 2;
	}
}