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)

#pragma GCC optimize("Ofast")
//#pragma comment(linker, "/STACK: 2000000")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include <bits/stdc++.h>
 
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define endl '\n'
#define elif else if
#define lli long long int
#define ld long double
#define vec vector
#define forn(a) for(lli a=0; a<n; a++)
#define all(x) x.begin(), x.end()
#define presicion(x) cout<<fixed<<setprecision(x)
#define deb(x) cerr<<#x<<" = "<<x<<endl
#define PI 3.14159265358979323
#define sen(ang) sin((ang)*PI/180)
 
using namespace std;
 
int main(){
    IOS;
  
  	int a, b; cin>>a>>b;
  	if(b%a==0) cout<<(a+b)<<endl;
  	else cout<<(b-a)<<endl;
 
    return 0;
}