CodeForce Round #479 A. Karen and Morning

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A. Karen and Morning

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, because05:39 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.

题意是给你一个时间，问多少分钟后，这个时间能成为回文时间。由于时间较少，所以可以直接枚举。

代码实现：

#include<iostream>#include<algorithm>using namespace std;int main(){char a[15];int i;while(cin>>a){int h=(a[0]-'0')*10+(a[1]-'0');int m=(a[1]-'0')*10+(a[0]-'0');int mm=(a[3]-'0')*10+(a[4]-'0');if(mm==m){cout<<"0"<<endl;continue;}for(i=0;;i++){if(mm==m)break;mm++;if(mm==60){h++;h=h%24;mm=0;m=(h%10)*10+h/10;}}cout<<i<<endl;}return 0;}