CodeForces - 916A

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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.

Example

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：有一个人特别爱睡，他定闹钟呢，有个毛病，就是7作为他的幸运数，闹钟必须从带有7这个数字的点开始响，问到他起床的时候 闹钟已经响了多少次，闹钟每隔x分钟响一次。

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分析：

从他起床的时间开始往前减x分钟，直到遇见这个幸运的时间。没什么难度，注意时间的表示。

#include <iostream>
#include <stdio.h>
#include <stack>
#include <stdlib.h>
#include <malloc.h>
#include <cmath>
#include <algorithm>
using namespace std;
bool Is(int qq,int ww)
{
    if(qq==7||qq%10==7||ww==7||ww%10==7)
    {
        return true;
    }
    else
        return false;
}
int main()
{
    int hh,mm;
    int x;
    cin>>x;
    cin>>hh>>mm;
    int sum=0;
    if(hh==7||hh%10==7||mm==7||mm%10==7)
    {
        cout<<0<<endl;
        return 0;
    }
    else
    {
        while(1)
        {
            if(mm-x<0)
            {
                mm=60-x+mm;
                hh--;
                if(hh<0)
                {
                    hh=23;
                }
            }
            //if(mm-x==0)

            else
            {
                mm-=x;
            }
            //cout<<hh<<" "<<mm<<endl;
            sum++;
            if(Is(hh,mm)==true)
                break;

        }
    }
    cout<<sum<<endl;
}