Problem:
Score: 600 points

Problem Statement

Takahashi and Aoki are training for long-distance races in an infinitely long straight course running from west to east.

They start simultaneously at the same point and moves as follows towards the east:

Takahashi runs A_1 meters per minute for the first T_1 minutes, then runs at A_2 meters per minute for the subsequent T_2 minutes, and alternates between these two modes forever.

Aoki runs B_1 meters per minute for the first T_1 minutes, then runs at B_2 meters per minute for the subsequent T_2 minutes, and alternates between these two modes forever.

How many times will Takahashi and Aoki meet each other, that is, come to the same point? We do not count the start of the run. If they meet infinitely many times, report that fact.

Constraints

1 \leq T_i \leq 100000

1 \leq A_i \leq 10^{10}

1 \leq B_i \leq 10^{10}

A_1 \neq B_1

A_2 \neq B_2

All values in input are integers.

Input

Input is given from Standard Input in the following format:

T_1 T_2
A_1 A_2
B_1 B_2

Output

Print the number of times Takahashi and Aoki will meet each other.

If they meet infinitely many times, print infinity instead.

Sample Input 1

1 2
10 10
12 4

Sample Output 1

1

They will meet just once, \frac{4}{3} minutes after they start, at \frac{40}{3} meters from where they start.

Sample Input 2

100 1
101 101
102 1

Sample Output 2

infinity

They will meet 101, 202, 303, 404, 505, 606, ... minutes after they start, that is, they will meet infinitely many times.

Sample Input 3

12000 15700
3390000000 3810000000
5550000000 2130000000

Sample Output 3

113

The values in input may not fit into a 32-bit integer type.

Solution in C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Text.RegularExpressions;

public class Program
{
    private long T1, T2;
    private long A1, A2;
    private long B1, B2;

    public void Solve()
    {
        var sc = new Scanner();
        T1 = sc.NextInt();
        T2 = sc.NextInt();
        A1 = sc.NextLong() * T1;
        A2 = sc.NextLong() * T2;
        B1 = sc.NextLong() * T1;
        B2 = sc.NextLong() * T2;

        long aa = A1 + A2;
        long bb = B1 + B2;
        if (aa == bb)
        {
            Console.WriteLine("infinity");
            return;
        }

        if (aa > bb)
        {
            Swap(ref aa, ref bb);
            Swap(ref A1, ref B1);
            Swap(ref A2, ref B2);
        }

        // aa < bb

        if (B1 > A1)
        {
            Console.WriteLine("0");
            return;
        }

        // B1 < A1

        long ans = 0;

        // T1で追い付く
        {
            // T1 開始時 AがBに追いつける位置にいる
            long d = A1 - B1;
            ans += (d) / (bb - aa);
        }

        // T2
        {
            // T2開始時 Aが前にいる
            // T1開始時 AがBに追いつける位置にいる = T1の条件
            ans *= 2;
        }

        Console.WriteLine(ans + 1);
    }

    void Swap(ref long x, ref long y)
    {
        long t = x;
        x = y;
        y = t;
    }

    public static void Main(string[] args)
    {
        new Program().Solve();
    }
}

class Scanner
{
    public Scanner()
    {
        _pos = 0;
        _line = new string[0];
    }

    const char Separator = ' ';
    private int _pos;
    private string[] _line;

    #region スペース区切りで取得

    public string Next()
    {
        if (_pos >= _line.Length)
        {
            _line = Console.ReadLine().Split(Separator);
            _pos = 0;
        }

        return _line[_pos++];
    }

    public int NextInt()
    {
        return int.Parse(Next());
    }

    public long NextLong()
    {
        return long.Parse(Next());
    }

    public double NextDouble()
    {
        return double.Parse(Next());
    }

    #endregion

    #region 型変換

    private int[] ToIntArray(string[] array)
    {
        var result = new int[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = int.Parse(array[i]);
        }

        return result;
    }

    private long[] ToLongArray(string[] array)
    {
        var result = new long[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = long.Parse(array[i]);
        }

        return result;
    }

    private double[] ToDoubleArray(string[] array)
    {
        var result = new double[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = double.Parse(array[i]);
        }

        return result;
    }

    #endregion

    #region 配列取得

    public string[] Array()
    {
        if (_pos >= _line.Length)
            _line = Console.ReadLine().Split(Separator);

        _pos = _line.Length;
        return _line;
    }

    public int[] IntArray()
    {
        return ToIntArray(Array());
    }

    public long[] LongArray()
    {
        return ToLongArray(Array());
    }

    public double[] DoubleArray()
    {
        return ToDoubleArray(Array());
    }

    #endregion
}