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>
#include <ctype.h>
using namespace std;

#define ll long long
#define all(x) (x).begin(), (x).end()
const long long INF = 1LL << 60;

#define rep(i, n) for (int i = 0; i < (int)(n); i++)  //(i, 10) i=0～i=9まで
#define repr(i, n) for (int i = n; i >= 0; i--)   // (i, 10) i=10～i=0まで
#define FOR(i, m, n) for (int i = m; i < n; i++)  // (i, 3, 10) i=3～i=9まで

// 総数を1000000007で割った余り
const long long mod = 1e9 + 7;

int main() {
    // cin.tie(0);
    // ios::sync_with_stdio(false);
    // cout << fixed << setprecision(5);

    // 入力
    int n;
    cin >> n;

    // 処理
    if (n % 2 == 0) {
        cout << n << endl;

    } else {
        cout << n * 2 << endl;
    }

    return 0;
}