A. Karen and Morning

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, 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.

 题解：
一个一个往前面推，每次判断时间是否回文就可以了。

#include<iostream>
#include<cstdio> 
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
using namespace std;
int cnt;
bool judge(int a,int b)
{
    if(a/10==b%10&&a%10==b/10)return 1;
    else return 0;
}
int main()
{
    int a,b;
    scanf("%d:%d",&a,&b);
    if(judge(a,b)){cout<<0;return 0;}
    while(!judge(a,b))
    {
        b++;
        cnt++;
        if(b==60){a++;b=0;}
        if(a==24){a=0;}
    }
    cout<<cnt;
    return 0;
}