CF-845B

B. Luba And The Ticket

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Luba has a ticket consisting of 6 digits. In one move she can choose digit in any position and replace it with arbitrary digit. She wants to know the minimum number of digits she needs to replace in order to make the ticket lucky.

The ticket is considered lucky if the sum of first three digits equals to the sum of last three digits.

Input

You are given a string consisting of 6 characters (all characters are digits from 0 to 9) — this string denotes Luba's ticket. The ticket can start with the digit 0.

Output

Print one number — the minimum possible number of digits Luba needs to replace to make the ticket lucky.

Examples

input

000000

output

0

input

123456

output

2

input

111000

output

1

Note

In the first example the ticket is already lucky, so the answer is 0.

In the second example Luba can replace 4 and 5 with zeroes, and the ticket will become lucky. It's easy to see that at least two replacements are required.

In the third example Luba can replace any zero with 3. It's easy to see that at least one replacement is required.

 

 

 

题意：

对于给出的长度为6的整数串，求最小改变多少个整数可以使前三个数之和等于后三个数。

 

起初直接暴力写太复杂了，可以先求出前后的和s1，s2 。将较小的一边三个数变为9-a[i]，即将它们全部变为9时差值的变化量，而后排序，从大到小减差值直到差值不大于0即可。

 

AC代码：

#include<bits/stdc++.h>
using namespace std;

int a[10];

int main(){
    ios::sync_with_stdio(false);
    string s;
    cin>>s;
    int s1=0,s2=0;
    for(int i=0;i<3;i++){
        a[i]=s[i]-'0';
        s1+=a[i];
    }
    for(int i=3;i<6;i++){
        a[i]=s[i]-'0';
        s2+=a[i];
    }
    int num=s1-s2; 
    if(num>0){
        for(int i=3;i<6;i++){
            a[i]=9-a[i];
        }
    }
    else{
        for(int i=0;i<3;i++){
            a[i]=9-a[i];
        }
        num=-num;
    }
    sort(a,a+6);
    for(int i=5;i>0;i--){
        if(num<=0){
            cout<<5-i<<endl;
            break;
        }
        num-=a[i];
    }
    return 0;
}