题目连接：

http://www.codeforces.com/contest/665/problem/A

Description

Buses run between the cities A and B, the first one is at 05:00 AM and the last one departs not later than at 11:59 PM. A bus from the city A departs every a minutes and arrives to the city B in a ta minutes, and a bus from the city B departs every b minutes and arrives to the city A in a tb minutes.

The driver Simion wants to make his job diverse, so he counts the buses going towards him. Simion doesn't count the buses he meet at the start and finish.

You know the time when Simion departed from the city A to the city B. Calculate the number of buses Simion will meet to be sure in his counting.

Input

The first line contains two integers a, ta (1 ≤ a, ta ≤ 120) — the frequency of the buses from the city A to the city B and the travel time. Both values are given in minutes.

The second line contains two integers b, tb (1 ≤ b, tb ≤ 120) — the frequency of the buses from the city B to the city A and the travel time. Both values are given in minutes.

The last line contains the departure time of Simion from the city A in the format hh:mm. It is guaranteed that there are a bus from the city A at that time. Note that the hours and the minutes are given with exactly two digits.

Output

Print the only integer z — the number of buses Simion will meet on the way. Note that you should not count the encounters in cities A and B.

Sample Input

10 30
10 35
05:20

Sample Output

5

Hint

题意

有两个城市A,B。

每a分钟A会发一趟车，这辆车会走TA秒到达B。

每b分钟B会发一趟车，这辆车会走TB秒到达A。

两个城市从早上5点开始发车，11点停止发车

有一个小朋友，坐着从A到B的列车，问他能够看到多少辆车

题解：

看到的车显然就是在这辆车行进的这个时间段内，B地才开始发车的和没有到的那些车。

然后暴力一波就好了~

代码

#include<bits/stdc++.h>
using namespace std;
int main()
{int a,ta,b,tb;scanf("%d%d%d%d",&a,&ta,&b,&tb);int h,m;scanf("%d:%d",&h,&m);m+=60*h;int ans = 0;for(int i=300;i<24*60;i+=b)if(i+tb>m&&i<m+ta)ans++;cout<<ans<<endl;
}