using System;
using System.Collections.Generic;
using System.Linq;

class Program
{
    static string InputPattern = "InputX";

    static List<string> GetInputList()
    {
        var WillReturn = new List<string>();

        if (InputPattern == "Input1") {
            WillReturn.Add("3");
            WillReturn.Add("1 0 4 1");
            WillReturn.Add("2 4 15 4");
            WillReturn.Add("0 0 1000000000000000000 1000000000000000000");
            //3
            //0
            //1000000000000000000
        }
        else {
            string wkStr;
            while ((wkStr = Console.ReadLine()) != null) WillReturn.Add(wkStr);
        }
        return WillReturn;
    }

    static long[] GetSplitArr(string pStr)
    {
        return (pStr == "" ? new string[0] : pStr.Split(' ')).Select(pX => long.Parse(pX)).ToArray();
    }

    static void Main()
    {
        List<string> InputList = GetInputList();

        long[] wkArr = { };
        Action<string> SplitAct = (pStr) => wkArr = GetSplitArr(pStr);

        foreach (string EachStr in InputList.Skip(1)) {
            SplitAct(EachStr);
            long SX = wkArr[0];
            long SY = wkArr[1];
            long TX = wkArr[2];
            long TY = wkArr[3];
            long Answer = Solve(SX, SY, TX, TY);
            Console.WriteLine(Answer);
        }
    }

    static long Solve(long pSX, long pSY, long pTX, long pTY)
    {
        long Answer = 0;

        // Y座標の移動コスト
        Answer += Math.Abs(pSY - pTY);

        long AnswerY = Math.Max(pSY, pTY);
        long Beki2 = 1;
        for (long I = 1; I <= AnswerY; I++) {
            if (WillOverProd(Beki2, 2, pSX) && WillOverProd(Beki2, 2, pTX)) {
                Beki2 = long.MaxValue;
                break;
            }
            else {
                Beki2 *= 2;
            }
        }

        long Div1 = (pSX / Beki2);
        long Div2 = (pTX / Beki2);
        Answer += Math.Abs(Div1 - Div2);
        return Answer;
    }

    // long型の2正数の掛け算が、Limitを超えるかを返す
    static bool WillOverProd(long pA, long pB, long pLimit)
    {
        return pA > pLimit / pB;
    }
}