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>
int main()
{
  char s[5];
  int n[4], i;
  scanf("%s", &s[0]);
  for(i=0;i<4;i++) n[i] = s[i] - '0';
  if((n[0] + n[1] + n[2] + n[3] ) == 7) printf("%d+%d+%d+%d=7\n", n[0], n[1], n[2], n[3]);
  else if((n[0] + n[1] + n[2] - n[3] ) == 7) printf("%d+%d+%d-%d=7\n", n[0], n[1], n[2], n[3]);
  else if((n[0] + n[1] - n[2] + n[3] ) == 7) printf("%d+%d-%d+%d=7\n", n[0], n[1], n[2], n[3]);
  else if((n[0] - n[1] + n[2] + n[3] ) == 7) printf("%d-%d+%d+%d=7\n", n[0], n[1], n[2], n[3]);
  else if((n[0] + n[1] - n[2] - n[3] ) == 7) printf("%d+%d-%d-%d=7\n", n[0], n[1], n[2], n[3]);
  else if((n[0] - n[1] - n[2] + n[3] ) == 7) printf("%d-%d-%d+%d=7\n", n[0], n[1], n[2], n[3]);
  else if((n[0] - n[1] + n[2] - n[3] ) == 7) printf("%d-%d+%d-%d=7\n", n[0], n[1], n[2], n[3]);
  else if((n[0] - n[1] - n[2] - n[3] ) == 7) printf("%d-%d-%d-%d=7\n", n[0], n[1], n[2], n[3]);
  else printf("error\n");
  return 0;
}