Codeforces Round #457 (Div. 2) A. Jamie and Alarm Snooze简单模拟

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 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 notlucky.

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 alucky 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: 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.

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

模拟即可，注意小时与分钟间的切换，00:00的前时刻为23:59

#include 
using namespace std;
int main()
{
    int x,hh,mm;
    while (~scanf("%d %d %d",&x,&hh,&mm)){
        int f=0,ans=0;
        while (1){
            if ((hh==17)||(hh==7)||(mm%10==7))
                break;
            ans++;
            mm-=x;
            if (mm<0){
                mm+=60;
                hh--;
            }
            if (hh<0)
                hh=23;
        }
        printf("%d\n",ans);
    }
    return 0;
}