The 9-th grade student Gabriel noticed a caterpillar on a tree when walking around in a forest after the classes. The caterpillar was on the height h1 cm from the ground. On the height h2 cm (h2 > h1) on the same tree hung an apple and the caterpillar was crawling to the apple.

Gabriel is interested when the caterpillar gets the apple. He noted that the caterpillar goes up by a cm per hour by day and slips down by b cm per hour by night.

In how many days Gabriel should return to the forest to see the caterpillar get the apple. You can consider that the day starts at 10 am and finishes at 10 pm. Gabriel’s classes finish at 2 pm. You can consider that Gabriel noticed the caterpillar just after the classes at 2 pm.

Note that the forest is magic so the caterpillar can slip down under the ground and then lift to the apple.

Input
The first line contains two integers h1, h2 (1 ≤ h1 < h2 ≤ 105) — the heights of the position of the caterpillar and the apple in centimeters.

The second line contains two integers a, b (1 ≤ a, b ≤ 105) — the distance the caterpillar goes up by day and slips down by night, in centimeters per hour.

Output
Print the only integer k — the number of days Gabriel should wait to return to the forest and see the caterpillar getting the apple.

If the caterpillar can’t get the apple print the only integer  - 1.

Example
Input
10 30
2 1
Output
1
Input
10 13
1 1
Output
0
Input
10 19
1 2
Output
-1
Input
1 50
5 4
Output
1
Note
In the first example at 10 pm of the first day the caterpillar gets the height 26. At 10 am of the next day it slips down to the height 14. And finally at 6 pm of the same day the caterpillar gets the apple.

Note that in the last example the caterpillar was slipping down under the ground and getting the apple on the next day.

題意
有一個人，白天每小時爬a米，晚上每小時掉下來b米，可以掉到地下去

你從第0天的下午兩點看到這個人在h1，他要到h2去

早上10點到晚上十點算白天，其他算晚上

問你他第幾天到達h2

如果不能到達輸出-1

#include<stdio.h>
using namespace std;
long long h1,h2,a,b;
int main()
{
    int now=14,day=0,t=0;
    scanf("%ld%ld%ld%ld",&h1,&h2,&a,&b);
    while(t<1e7)
    {
        t++;
        if(now>=10&&now<=21) h1+=a;
        else h1-=b;
        if(h1>=h2)
        {
            printf("%ld\n",day);
            return 0;
        }
        now=(now+1)%24;
        if(now==0) day++;
    }
    printf("-1\n");
    return 0;
}