916A - Jamie and Alarm Snooze

A. Jamie and Alarm Snooze

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Jamie loves sleeping. One day, he decides that he needs to wake up at exactly hh: mm. However, he hates waking up, so he wants
to make waking up less painful by setting the alarm at a luckytime. He will then press the snooze button every x minutes
until hh: mm is reached, and only then he will wake up. He wants to know what is the smallest number of times he needs to
press the snooze button.

A time is considered lucky if it contains a digit '7'.
For example, 13: 07 and 17: 27 are lucky,
while 00: 48 and 21: 34 are
not lucky.

Note that it is not necessary that the time set for the alarm and the wake-up time are on the same day. It is guaranteed that there is a lucky time Jamie can
set so that he can wake at hh: mm.

Formally, find the smallest possible non-negative integer y such that the time representation of the time x·y minutes
before hh: mm contains the digit '7'.

Jamie uses 24-hours clock, so after 23: 59 comes 00: 00.

Input

The first line contains a single integer x (1 ≤ x ≤ 60).

The second line contains two two-digit integers, hh and mm (00 ≤ hh ≤ 23, 00 ≤ mm ≤ 59).

Output

Print the minimum number of times he needs to press the button.

Examples

input

3
11 23

output

2

input

5
01 07

output

0

Note

In the first sample, Jamie needs to wake up at 11:23. So, he can set his alarm at 11:17. He would press the snooze button when the alarm rings at 11:17 and at 11:20.

In the second sample, Jamie can set his alarm at exactly at 01:07 which is lucky.

题意：一个人喜欢睡觉，他在设置某个时间起床，如果不是幸运数字的话，如果h或者m不带有7，就会每隔x分钟响一次。他不想被打扰，所i以不是幸运数的时间，就要按一次闹钟，才能继续睡。所以要求他定的闹钟时间和前面最近的的幸运数的时间，需要按的次数，就是最少需要按多少次闹钟。

题解：模拟  一,如果h或者m带有7，不需要按，直接输出0。其余就用当前的m减去x，次数++。不够就时间的进制。

#include<bits/stdc++.h>
using namespace std;
int main()
{
int x,h,m,ans;
while(cin>>x)
{
ans=0;
cin>>h>>m;
if(m%10==7||h%10==7)
puts("0");
else
{
while(m%10!=7&&h%10!=7)
{
m-=x;
ans++;
if(m<0)
{
m+=60,h--;
if(h<0)
h+=24;
}
}
cout<<ans<<endl;
}
return 0;
}
}