由1变成2，按红蓝按钮皆可，所以特殊处理1

由1变成2，按红蓝按钮皆可，所以特殊处理1；
从A变成a，从B变成b，只要是偶数就是蓝色按钮的逆操作。（底数为2的对数复杂度，步数最少）

import java.util.*;
public class Main {
    public static void main(String[] args) throws Exception{
        Scanner sc = new Scanner(System.in);
        int a = sc.nextInt(), b = sc.nextInt();
        int A = sc.nextInt(), B = sc.nextInt();
        StringBuilder button1 = new StringBuilder("");    //A变成a的按钮序列
        while(A > 0 && A != a){
            if(A%2 == 0){
                button1.append('B');
                A /= 2;
            }else{
                button1.append('R');
                A--;
            }
        }
        StringBuilder button2 = new StringBuilder("");    //B变成b的按钮序列
        while(B > 0 && B != b){
            if(B%2 == 0){
                button2.append('B');
                B /= 2;
            }else{
                button2.append('R');
                B--;
            }
        }
        if(button1.toString().equals(button2.toString())){
            System.out.println(button1.length());
        }else{
            int n = button1.length();
            int m = button2.length();
            if((a == 1 || b == 1)&&button1.toString().substring(0, n - 1).equals(button2.toString().substring(0, m - 1))){
                System.out.println(button1.length());
            }
            else System.out.println(-1);
        }
    }
}