Problem:
Score : 100 points

Problem Statement

You are given positive integers A and B.

If A is a divisor of B, print A + B; otherwise, print B - A.

Constraints

All values in input are integers.

1 \leq A \leq B \leq 20

Input

Input is given from Standard Input in the following format:

A B

Output

If A is a divisor of B, print A + B; otherwise, print B - A.

Sample Input 1

4 12

Sample Output 1

16

As 4 is a divisor of 12, 4 + 12 = 16 should be printed.

Sample Input 2

8 20

Sample Output 2

12

Sample Input 3

1 1

Sample Output 3

2

1 is a divisor of 1.

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

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;
using vs = vector<string>;
using pll = pair<ll, ll>;
using vp = vector<pll>;
#define rep(i, n) for (int i = 0; i < (n); i++)
#define repb(i, n) for (int i = (n-1); i >= 0; i--)
#define rea(i, a, b) for (int i = (a); i < (b); i++)
#define reab(i, a, b) for (int i = (a)-1; i >= (b); i--)
#define ALL(a) (a).begin(),(a).end()
#define SZ(x) ((int)(x).size())
#define PRINT(V) for (auto v:(V)) {cerr << v << " ";} cerr << endl
#define dump(x) cerr << #x << " : " << (x) << endl
#define dump2(x, y) cerr << "(" << #x << ", " << #y << ") = (" << x << ", " << y << ")" << endl
inline ll GCD(ll a, ll b){return b ? GCD(b, a % b) : a;}
const int MOD = 1000000007;
/* END OF TEMPLATE */

int main()
{
    ll a, b;
    cin >> a >> b;
    if(b % a == 0) cout << a + b << endl;
    else cout << b - a << endl;
}