Jamie and Alarm Snooze（起床时间，延迟闹铃）

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宇宙第一小仙女\(^o^)/～～萌量爆表求带飞=≡Σ((( つ^o^)つ~dalao们点个关注呗～～

emmm都喊我铁头，哼！！！而我的床呢偏偏就是磁铁打的........but每天还是要起床.....比如更博，therefore....没事多来瞅瞅给我增点访客量咯～这个叫Jamie的小盆友还真是像我一样爱睡觉，这个闹铃延迟提醒当然也是很常见了，再来一分钟，一分钟之后就起.......嚯！还非得在带“7”的时候设闹钟。

不扯了不扯了，hint 一下咯～ 这题倒是不算难，但是很容易WA，特殊情况比较多，多试试特殊情况就差不多找到错了，比如23点之后没有24点，要变成0点，时分秒到60要进1。

Problem is coming ~~~~

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 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.

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.

思路应该都有，就不发了，建议看代码之前还是自己先想想思路啦～

----------------------------------------------------------------我只是一条可爱的分界线---------------------------------------------------------------

#include<stdio.h>
int main()
{
    int k=0,x,h,m,i,j;
    scanf("%d",&x);
    scanf("%d%d",&h,&m);
    i=h;
    for(j=m;;j-=x){
        if(j<0){
            i--;
            j+=60;
        }
        if(i<0)
            i+=24;
        if(i%10==7||j%10==7){
            break;
        }
        k++;
    }
    printf("%d\n",k);
    return 0;
}

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