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: Wrong Answer)

//#include <bits/stdc++.h>
//#include <stdio.h>
#include<iostream>
#include<cstdio>
#include<bitset>
#include<algorithm>
#include<vector>
#include<list>
#include<queue>
#include<stack>
#include<string>
#include<string.h>
#include<cmath>
#include<utility>
#include<functional>
#include<map>
#include<set>
#include<cctype>
#include<fstream>


#define FOR(i, a, b) for( int i=(a);i<=(b);i++)
#define RFOR(i, a, b) for( int i=(a);i>=(b);i--)
#define LFOR(i, a, b) for( long long int i=(a);i<=(b);i++)
#define LRFOR(i, a, b) for(long long int i=(a);i>=(b);i--)
#define MOD 1000000007
#define INF 1000000000 //2000000000
#define LLINF 1000000000000000000 //9000000000000000000
#define PI 3.14159265358979
#define MAXI 7500000

using namespace std;
typedef long long int ll;
typedef pair< long long int, long long  int> P;
typedef string::const_iterator State;

int dy[5] = { 0,0,1,-1,0 };
int dx[5] = { 1,-1,0,0 ,0 };


int main(void) {
	
	int n;

	cin >> n;

	if (n % 2 == 0) {
		cout << 2 << endl;
	}
	else {
		cout << n * 2 << endl;
	}

	return 0;
}