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 Ruby (2.3.3) (judge verdict: Wrong Answer)

a, b, c, d = gets.chomp.split("").map(&:to_i)
op = []
flag = false
(0..1).each do |i|
    (0..1).each do |j|
        (0..1).each do |k|
            (0..1).each do |l|
                if (i*2-1)*a + (j*2-1)*b + (k*2-1)*c + (l*2-1)*d == 7
                    op.push(i, j, k, l)
                    flag = true
                end
                break if flag     
            end
            break if flag
        end
        break if flag
    end
    break if flag
end
(0..3).each do |i|
    if op[i] == 1
        op[i] = '+'
    else 
        op[i] = '-'
    end
end

puts "#{a}#{op[0]}#{b}#{op[1]}#{c}#{op[2]}#{d}#{op[3]}=7"