Codeforce916A

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<cstdio>
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
");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
",ans);
    return 0;
}