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 Rust (1.15.1) (judge verdict: Wrong Answer)

use std::io::*;
use std::str::*;

fn read<T: FromStr>() -> Option<T> {
    let stdin = stdin();
    let s = stdin
        .bytes()
        .map(|c| c.unwrap() as char)
        .take_while(|c| !c.is_whitespace())
        .collect::<String>();
    s.parse::<T>().ok()
}

fn main() {
    let s: String = read().unwrap();

    let a = s.chars()
        .nth(0)
        .unwrap()
        .to_string()
        .parse::<i32>()
        .unwrap();
    let b = s.chars()
        .nth(1)
        .unwrap()
        .to_string()
        .parse::<i32>()
        .unwrap();
    let c = s.chars()
        .nth(2)
        .unwrap()
        .to_string()
        .parse::<i32>()
        .unwrap();
    let d = s.chars()
        .nth(3)
        .unwrap()
        .to_string()
        .parse::<i32>()
        .unwrap();

    let mut op = vec!['+', '-'];

    for i in 0..op.len() {
        let mut ans: i32 = 0;
        if op[i] == '+' {
            ans = a + b;
        } else {
            ans = a - b;
        }
        for j in 0..op.len() {
            if op[j] == '+' {
                ans = ans + c;
            } else {
                ans = ans - c;
            }
            for k in 0..op.len() {
                if op[k] == '+' {
                    ans = ans + d;
                } else {
                    ans = ans - d;
                }
                if ans == 7 {

                    println!("{}{}{}{}{}{}{}=7", a, op[i], b, op[j], c, op[k], d);
                    return;
                }
            }
        }
    }
    return;
}