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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using ll = long long;

#define mod 1000000007
#define REP(i, n) for (int i = 0; i < n; i++)
#define INF (1 << 29)

using namespace std;

//long C(long n, long r);
//long long gcd(long long a, long long b);
//long long lcm(long long a, long long b);


void solve()
{
    string t;
    cin >> t;
    char opt[2] = {'+', '-'};
    for(int i = 0; i <= 1; i++){
        for(int j = 0; j <= 1; j++){
            for(int k = 0; k <= 1; k++) {
                if((t[0]-'0') + (int)pow(-1, i) * (t[1]-'0') + (int)pow(-1, j) * (t[2]-'0') + (int)pow(-1, k) * (t[3]-'0') == 7){
                    cout << t[0]-'0' << opt[i] << t[1]-'0' << opt[j] << t[2]-'0' << opt[k] << t[3]-'0' << "=7" << endl;
                    return;
                }
            }
        }
    }
}

int main()
{
    solve();
    return 0;
}