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 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rrep(i, n) for (int i = 1; i < (int)(n+1); i++)
//二項係数の計算
int nCr(int A, int B) {
    ll ans = 1;
    for (int i = 0; i < B; i++) {
        ans *= A - i;
    }
    for (int i = 0; i < B; i++) {
        ans /= i + 1;
    }
    return ans;
}
// vector<vector<int>> A(3, vector<int>(4));
int main() {
    char F,G,H,I;
  cin>>F>>G>>H>>I;
  int A='0'+F-96;
  int B='0'+G-96;
  int C='0'+H-96;
  int D='0'+I-96;
    vector<char>E(2);
    E={'+','-'};
    rep(i,2){
        rep(j,2){
            rep(k,2){
                int cnt=A;
                if(E[i]=='+'){
                    cnt+=B;
                }
                else{
                    cnt-=B;
                }
                if(E[j]=='+'){
                    cnt+=C;
                }
                else{
                    cnt-=C;
                }
                if(E[k]=='+'){
                    cnt+=D;
                }
                else{
                    cnt-=D;
                }
                if(cnt==7){
                    cout<<A<<E[i]<<B<<E[j]<<C<<E[k]<<D<<"="<<7<<endl;
                	break;
                }
            }
        }
    }
}