#include <iostream>
#include <string>
#include <string.h>
#include <stdio.h>
#include <math.h>
#include <algorithm>

using namespace std;

int quick_pow(int base, int power)
{
    int result = 1;
    while(power)
    {
        if(power & 1)
        {
            result = result * base;//如果求后三位 那就mod1000
        }
        power >>= 1;
        base = base * base;//如果求后三位 那就mod1000
    }
    return result;
}

int main()
{
    string s; getline(cin, s);
    int ls = s.size();

    int vis = -1; bool f1 = 0; bool f2 = 0;

    if(s[0] == '-' || s[0] == '0') f1 = 1;//负数

    for(int i = 0; i < ls; i++)
    {
        if(s[i] == ' ')
        {
            vis = i; break;
        }

        if(s[i] == '.') f1 = 1;//小数
        else if(s[i] < '0' || s[i] > '9') f1 = 1;//乱码
        else if(i > 3) f1 = 1;
        else if(i == 3 && s[0] > '1') f1 = 1;
        else if(i == 3 && s[0] == '1' && (s[1] != '0' || s[2] != '0' || s[3] != '0')) f1 = 1;

    }
    if(vis == 0)
    {
        f1 = 1;
        vis++;
    }


    if(s[vis + 1] == '-' || s[vis + 1] == '0') f2 = 1;
    for(int i = vis + 1, j = 0; i < ls; i++, j++)
    {
        if(s[i] == '.') f2 = 1;
        else if(s[i] < '0' || s[i] > '9') f2 = 1;//乱码
        else if(j > 3) f2 = 1;
        else if(j == 3 && s[vis + 1] > '1') f2 = 1;
        else if(j == 3 && s[vis + 1] == '1' && (s[vis + 2] != '0' || s[vis + 3] != '0' || s[vis + 4] != '0')) f2 = 1;
    }

    if(f1 && f2) cout<<"? + ? = ?\n";
    else if(f1)
    {
        cout<<"? + ";
        for(int i = vis + 1; i < ls; i++) cout<<s[i];
        cout<<" = ?\n";
    }
    else if(f2)
    {
        for(int i = 0; i < vis; i++) cout<<s[i];
        cout<<" + ? = ?\n";
    }
    else
    {
        int na = 0; int nb = 0;
        for(int i = vis - 1, j = 0; i >= 0; i--, j++)
        {
            int t = s[i] - '0';
            na += t * quick_pow(10, j);
        }
        for(int i = ls - 1, j = 0; i >= vis + 1; i--, j++)
        {
            int t = s[i] - '0';
            nb += t * quick_pow(10, j);
        }
        cout<<na<<" + "<<nb<<" = "<<na + nb<<'\n';
    }

    return 0;
}