A. Karen and Morning(Round419)

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.

 

 hint:给你一个时间，让你求这个时间到下一个回文时间需要多久。。。

这个题目，我刚开始还真是想着去把所有的回文数字给找出来，然后挨个判断，结果可想而知啊。。吃力不讨好。。。

看完题解之后，我发现。。。是不是div2的前几题都是可以大力出奇迹的题。。。

所以这个题目就是大力搞一发好了。。。将所给时间加1加1这样迭代加下去直到它成为一个回文数字。。。因为一天也就是24*60*3600s 。。。怎么暴力都不会超时的。。。

#include<iostream>
#include<cstdio>
#include<stack>
#include<queue>
#include<vector>
#include<cmath>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
using namespace std;
#define pb push_back
#define mk make_pair
#define FOR(i, a, b) for(int i=a; i<=b; i++)
#define FO(i, a, b) for(int i=a; i<b; i++)
#define MOD 1000000007
#define pi acos(-1.0)
#define eps 1e-8
#define mem(a, b) memset(a, b, sizeof(a))
typedef long long ll;
typedef pair<int, int> PII;
typedef vector<int> VI;

int main(){
    int hh, mm;
    scanf("%d:%d", &hh, &mm);
    int cnt = 0;
    while(1){
        if(hh%10*10+hh/10 == mm) break;
        if(mm == 59){   　　　　　　　　//在几个特殊的地方要判断一下，比如分钟为59的时候，小时为23的时候。。。
            cnt++;
            mm = 0;
            if(hh != 23) hh++;
            else hh = 0;
        }else{
            mm++;
            cnt++;
        }
    }
    printf("%d\n", cnt);
    return 0;
}