Problem:
Score : 300 points

Problem Statement

Sitting in a station waiting room, Joisino is gazing at her train ticket.

The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive).

In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with + or - so that the formula holds.

The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.

Constraints

0≤A,B,C,D≤9

All input values are integers.

It is guaranteed that there is a solution.

Input

Input is given from Standard Input in the following format:

ABCD

Output

Print the formula you made, including the part =7.

Use the signs + and -.

Do not print a space between a digit and a sign.

Sample Input 1

1222

Sample Output 1

1+2+2+2=7

This is the only valid solution.

Sample Input 2

0290

Sample Output 2

0-2+9+0=7

0 - 2 + 9 - 0 = 7 is also a valid solution.

Sample Input 3

3242

Sample Output 3

3+2+4-2=7

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

#include <stdio.h>
#include <stdlib.h>
int main()
{
   int a,b,c,d,num;
   while(scanf("%d",&num)!=EOF)
   {
       d=num%10;
       c=num/10%10;
       b=num/100%10;
       a=num/1000;
       int op1,op2,op3,e1,e2,f1,f2,g,y1,y2,y3;
       char x[3]={"+-"};
       for(op1=0;op1<2;op1++)
       {
            if(op1==0)
            {
                e1=a+b;
                for(op2=0;op2<2;op2++)
                {
                    if(op2==0)
                    {
                        f1=e1+c;
                        for(op3=0;op3<2;op3++)
                        {
                            if(op3==0) {g=f1+d;if(g==7) {y1=op1;y2=op2;y3=op3;}}
                            if(op3==1) {g=f1-d;if(g==7) {y1=op1;y2=op2;y3=op3;}}
                        }
                    }
                    if(op2==1)
                    {
                        f2=e1-c;
                        for(op3=0;op3<2;op3++)
                        {
                            if(op3==0)  {g=f2+d;if(g==7) {y1=op1;y2=op2;y3=op3;}}
                            if(op3==1)  {g=f2-d;if(g==7) {y1=op1;y2=op2;y3=op3;}}
                        }
                    }
                }
            }
            if(op1==1)
            {
                e2=a-b;
                for(op2=0;op2<2;op2++)
                {
                        if(op2==0)
                        {
                            f1=e2+c;
                            for(op3=0;op3<2;op3++)
                            {
                                if(op3==0) {g=f1+d;if(g==7) {y1=op1;y2=op2;y3=op3;}}
                                if(op3==1) {g=f1-d;if(g==7) {y1=op1;y2=op2;y3=op3;}}
                            }
                        }
                        if(op2==1)
                        {
                            f2=e2-c;
                            for(op3=0;op3<2;op3++)
                            {
                                if(op3==0) {g=f2+d;if(g==7) {y1=op1;y2=op2;y3=op3;}}
                                if(op3==1) {g=f2-d;if(g==7) {y1=op1;y2=op2;y3=op3;}}
                            }
                        }
                }
            }

       }
       if(g=7) printf("%d%c%d%c%d%c%d=7\n",a,x[y1],b,x[y2],c,x[y3],d);

   }
    return 0;
}