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.
import sys
import bisect as b
import math
from collections import defaultdict as dd
input=sys.stdin.readline
#sys.setrecursionlimit(10**7)
def cin():
    return map(int,sin().split())
def ain():            
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input())
##n=m=0
##s=''
##t=''
##dp=[]
##def solve(inds,indt,k,cont):
##    ans=-999999999999999
##    print(dp)
##    if(k<0):return 0
##    elif(inds>=n and indt>=m):return 0
##    elif(dp[inds][indt][k][cont]!=-1):return dp[inds][indt][k][cont]
##    else:
##        if(indt<m):ans=max(ans,solve(inds,indt+1,k,0))
##        if(inds<n):ans=max(ans,solve(inds+1,indt,k,0))
##        if(s[inds]==t[indt]):
##            ans=max(ans,solve(inds+1,indt+1,k-1,1)+1)
##            if(cont):ans=max(ans,solve(inds+1,indt+1,k,1)+1)
##        dp[inds][indt][k][cont]=ans
##        return ans

    
for _ in range(1):
    k=inin()
    h,m=cin()
    c=0
    s=str(h)+':'+str(m)
    while('7' not in s):
        if(k>m):
            if(h>0):
                h-=1
                m=60-(k-m)
            else:
                h=23
                m=60-(k-m)
        else:
            m-=k
        s=str(h)+':'+str(m)
##        print(s)
        c+=1
    print(c)
        
        
    
    







    
##    n,m,k=cin()
##    s=sin().strip()
##    t=sin().strip()
##    dp=[[[[-1]*2 for i in range(k)] for i in range(m+1)] for i in range(n+1)]
##    c=0
##    for i in dp:
##        for j in i:
##            for l in j:
##                c+=1
##                print(l,c)
##    print(solve(0,0,k,0))
    
        
        
            
            
        
        
    
    
    
            
    
        
        
        
    
            
    
    
    
    
        
                
            
        
        
        
    
            
            
        
    
    
        
    

        
        
    
                
                
                
                

        
            
            
                
            
                
    
 
        
    
    
    
    




    


   
        
    
    
    
    




    


