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 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.

 

分析：水题。给出一个时间hh:mm(24小时制),每次可以把时间往后退x分钟，

直到hh:mm中出现数字7为止，输出后退次数。

 

#include
        
         int f(int a){    while(a)    {        if(a%10==7) return 1;        a/=10;    }    return 0;}int main(){    int x,a,b;    scanf("%d%d%d",&x,&a,&b);    if(f(a)||f(b)) {printf("0\n");return 0;}    int ans=0;    while(true)    {        if(b-x<0)//后退        {            b=60-(x-b);            if(a==0) a=23;            else a=a-1;        }        else b=b-x;        ans++;        if(f(a)||f(b)) break;    }    printf("%d\n",ans);    return 0;}