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 <iostream>
#include <string>
#include <vector>
#include <climits>
#include <set>
#include <algorithm>
#include <unordered_map>
#include <queue>
#include <iomanip>
#include <map>
#include <utility>
#include <numeric>
#include <chrono>
#include <ctime>
#include <bitset>
#include <iterator>
#include <cmath>
#include <stack>
#include <random>

#define rep(i,n) for(int i=0; i<(int)(n); i++)
#define debug(x) cerr << #x << ": " << x << ", "
#define debugln(x) cerr << #x << ": " << x << '\n'

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

using namespace std;
using ll = long long;
using P = pair<int, int>;

constexpr ll mod = 1e9+7;
const double PI = acos(-1.0);
mt19937_64 engine(chrono::steady_clock::now().time_since_epoch().count());

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);

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