Jamie and Alarm Snooze

Description

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 lucky time. 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 luckytime 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: mmcontains 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.

Sample Input

Input

3
11 23

Output

2

Input

5
01 07

Output

0

Hint

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.

题目意思：一个人想要在hh：mm时刻起床，而他想从在离起床时刻最近的一个带有数字7的时刻起，每x分钟按一次按钮，问一共按了多少次

解题思路：模拟即可，注意小时与分钟间的切换。

 #include <iostream>
 #include <stdio.h>
 using namespace std;
 int main()
 {
     int n,h,m,count;
     while(scanf("%d%d%d",&n,&h,&m)!=EOF)
     {
         count=;
         while()
         {
             if(h==||h==||m%==)
             {
                 break;
             }
             else
             {
                 count++;
                 m=m-n;
                 if(m<)
                 {
                     m=m+;
                     h--;
                 }
                 if(h<)
                 {
                     h=h+;
                 }
             }
         }
         printf("%d\n",count);
     }
     return ;
 }