A Plus Equals B

Problem:

A + B is a problem used to test one's basic knowledge for competitive programming. Here is yet another boring variation of it.

You have two integers, A and B. You want to make them equal. To do so, you can perform several steps, where each step is one of the following:

• A+=A
• A+=B
• B+=A
• B+=B

Unfortunately, A + B is a hard problem for us, so you are allowed to make at most 5000 steps.

Input

Two integers A, B are given. (1 ≤ A, B ≤ 10¹⁸).

Output

In the first line, print a single integer n (0 ≤ n ≤ 5000) denoting the number of steps.

In next n lines, print one of the following strings to denote your desired operation: "A+=A", "A+=B", "B+=A", or "B+=B".

Any sequence of steps that yields the desired result will be judged correct.

Example

Input

2 3

Output

4
B+=B
B+=A
A+=A
A+=A

    #include <iostream>
    #include <vector>
    using namespace std;
    #define int long long
    int a , b;
    vector < int > s;
    signed main ( void )
    {
        cin >> a >> b;
        while ( a != b ) {
            while ( a % 2 == 0 ) {
                s.push_back ( 1 );
                a /= 2;
            }
            while ( b % 2 == 0 ) {
                s.push_back ( 2 );
                b /= 2;
            }
            if ( a == b ) break;
            if ( a < b ) {
                b += a;
                s.push_back ( 3 );
            } else {
                a += b;
                s.push_back ( 4 );
            }
        }
        cout << s.size ( ) << endl;
        for ( int i = 0 ; i < s.size ( ) ; i++ ) {
            if ( s [ i ] == 1 ) {
                cout << "B+=B" << endl;
            } else if ( s [ i ] == 2 ) {
                cout << "A+=A" << endl;
            } else if ( s [ i ] == 3 ) {
                cout << "B+=A" << endl;
            } else {
                cout << "A+=B" << endl;
            }
        }
        return 0;
    }