Problem:
Score : 100 points

Problem Statement

You are given two integers a and b.
Determine if a+b=15 or a\times b=15 or neither holds.

Note that a+b=15 and a\times b=15 do not hold at the same time.

Constraints

1 \leq a,b \leq 15

All values in input are integers.

Input

Input is given from Standard Input in the following format:

a b

Output

If a+b=15, print +;
if a\times b=15, print *;
if neither holds, print x.

Sample Input 1

4 11

Sample Output 1

+

4+11=15.

Sample Input 2

3 5

Sample Output 2

*

3\times 5=15.

Sample Input 3

1 1

Sample Output 3

x

1+1=2 and 1\times 1=1, neither of which is 15.

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

#include<bits/stdc++.h> //Ithea Myse Valgulious
namespace chtholly{
typedef long long ll;
#define re0 register int
#define rec register char
#define rel register ll
#define gc getchar
#define pc putchar
#define p32 pc(' ')
#define pl puts("")
/*By Citrus*/
inline int read(){
  int x=0,f=1;char c=gc();
  for (;!isdigit(c);c=gc()) f^=c=='-';
  for (;isdigit(c);c=gc()) x=(x<<3)+(x<<1)+(c^'0');
  return f?x:-x;
  }
template <typename mitsuha>
inline bool read(mitsuha &x){
  x=0;int f=1;char c=gc();
  for (;!isdigit(c)&&~c;c=gc()) f^=c=='-';
  if (!~c) return 0;
  for (;isdigit(c);c=gc()) x=(x<<3)+(x<<1)+(c^'0');
  return x=f?x:-x,1;
  }
template <typename mitsuha>
inline int write(mitsuha x){
  if (!x) return 0&pc(48);
  if (x<0) x=-x,pc('-');
  int bit[20],i,p=0;
  for (;x;x/=10) bit[++p]=x%10;
  for (i=p;i;--i) pc(bit[i]+48);
  return 0;
  }
inline char fuhao(){
  char c=gc();
  for (;isspace(c);c=gc());
  return c;
  }
}using namespace chtholly;
using namespace std;

int main(){
int a=read(),b=read();
if (a+b==15) puts("+");
else if (a*b==15) puts("*");
else puts("x");
}