# Problem:

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.

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).

Print the minimum number of times he needs to press the button.

3 11 23

2

5 01 07

0

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.

# Solution:

# Note:

Solution is brute force. We know in the worst case, we could reach a time where the hh ends with a 7. The worst case is x = 2 and hh = 06 and mm = 58. This requires 13 * 30 + 29 = 419 times of subtraction in our code. This is definitely in the time limit.

Since 7 could only exist on the end of hour or minute, we only need to check for mod 10.