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 (1.42.0) (judge verdict: Accepted)

use whiteread::parse_line;

fn main() {
    let abcd: String = parse_line().unwrap();
    let mut abcd = abcd.chars();
    let a = abcd.next().unwrap().to_digit(10).unwrap() as i32;
    let b = abcd.next().unwrap().to_digit(10).unwrap() as i32;
    let c = abcd.next().unwrap().to_digit(10).unwrap() as i32;
    let d = abcd.next().unwrap().to_digit(10).unwrap() as i32;

    let mut op: Vec<i32> = vec![1, 1, 1];
    'outer: for x in 0..2 {
        let op_b;
        if x == 0 {
            op_b = 1;
        } else {
            op_b = -1;
        }
        for y in 0..2 {
            let op_c;
            if y == 0 {
                op_c = 1;
            } else {
                op_c = -1;
            }
            for z in 0..2 {
                let op_d;
                if z == 0 {
                    op_d = 1;
                } else {
                    op_d = -1;
                }
                if a + b * op_b + c * op_c + d * op_d == 7 {
                    op[0] = op_b;
                    op[1] = op_c;
                    op[2] = op_d;
                    break 'outer;
                }
            }
        }
    }
    let mut op = op.iter().map(|&x| if x == 1 { '+' } else { '-' });
    println!(
        "{}{}{}{}{}{}{}=7",
        a,
        op.next().unwrap(),
        b,
        op.next().unwrap(),
        c,
        op.next().unwrap(),
        d
    );
}