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 <cstdio>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <stdlib.h>
#include <string.h>
#include <cstring>
#include <cmath>
#include <map>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <stdio.h>
#include <tuple>
// *テンプレート*
#define int long long
#define rep(i, n) for (int i = 0; i < (n); i++)
#define P pair<long long,long long>
#define INF 1000000000000000000
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; }
const long long MOD = 1000000007;
using namespace std;
typedef long long ll;
map<int, int> mp;
using Graph = vector<vector<int>>; 
//最大公約数
ll gcd(ll a, ll b) {
	if (a < b)swap(a, b);
	if (b == 0) return a;
	return gcd(b, a % b);
}
//最小公倍数
ll lcm(ll a, ll b) {
	ll g = gcd(a, b);
	return a / g * b;
}

// *テンプレートここまで*



signed main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int a, b; cin >> a >> b;
	if (b % a == 0) cout << a + b << endl;
	else cout << b - a << endl;
	return 0;
}