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.
from __future__ import division
from math import ceil
from bisect import bisect_right as b_r
from bisect import bisect_left as b_l
t = input()
aa = raw_input()
if '7' in aa:
	print 0
else:
	x =  int(aa[3:])
	y = int(aa[:2])
	s = x
	i = 0
	cc= 0
	if y > 7 and y< 17:
		cc = y-7
	elif y<7:
		cc = 7+y
	elif y>7  and y <17:
		cc = 17-y
	else:
		cc = y-17
			
	if cc < 0:
		cc += 12
	mm =ceil((60*(cc-1)+x+1)*1.0/t)
	while(i<mm):
		s = s-t
		if s < 0:
			s = 60+s
		i+=1
		if '7' in str(s) :
			mm = i
			break
	print int(mm)
		
