codeforces 816A

time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

Karen is getting ready for a new school day!

It is currently hh:mm, given in a 24-hour format. As you know, Karen loves palindromes, and she believes that it is good luck to wake up when the time is a palindrome.

What is the minimum number of minutes she should sleep, such that, when she wakes up, the time is a palindrome?

Remember that a palindrome is a string that reads the same forwards and backwards. For instance, 05:39 is not a palindrome, because 05:39 backwards is 93:50. On the other hand, 05:50 is a palindrome, because 05:50 backwards is 05:50.

Input

The first and only line of input contains a single string in the format hh:mm (00 ≤  hh  ≤ 23, 00 ≤  mm  ≤ 59).

Output

Output a single integer on a line by itself, the minimum number of minutes she should sleep, such that, when she wakes up, the time is a palindrome.

Examples

input

05:39

output

11

input

13:31

output

0

input

23:59

output

1

Note

In the first test case, the minimum number of minutes Karen should sleep for is 11. She can wake up at 05:50, when the time is a palindrome.

In the second test case, Karen can wake up immediately, as the current time, 13:31, is already a palindrome.

In the third test case, the minimum number of minutes Karen should sleep for is 1 minute. She can wake up at 00:00, when the time is a palindrome.

题意：给你一个字符串，构成回文串，只是这个串数为时钟。

思路：简单模拟就可以了，注意特殊处理一些情况。

代码：

#include<bits/stdc++.h>using namespace std;const int maxn=10;int a[maxn];int main(){    string s;    int num1=0,num2=0;    while(cin>>s)    {        memset(a,0,sizeof(a));        int len=s.length();        int cnt=0;        for(int i=0;i<len;i++)        {            if(s[i]==':') continue;            a[cnt++]=s[i]-'0';        }        int num=a[0]*10+a[1];        num1=a[0]+a[1]*10;        num2=a[2]*10+a[3];        int sum=0;        if(num==23&&num2<=32)        {            sum=32-num2;            cout<<sum<<endl;            continue;        }        else if(num==23&&num2>32)        {            sum=60-num2;            cout<<sum<<endl;            continue;        }        else        {            while(1)            {                if(num2==60)                {                    num2=0;                   if(num1==90)                   {                       num1=1;                   }                   else if(num1==91)                   {                       num1=2;                   }                   else                   {                       num1+=10;                   }                }                if(num1==num2) break;                sum++;                num2++;            }            cout<<sum<<endl;        }    }    return 0;}