cfB. Luba And The Ticket time limit per test2 seconds memory limit per test256 megabytes inputstanda

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.

这题没能在2小时内做出来，但是也并不难。

开始的思路就是找规律，如果暴力求解，问题会复杂化。

感觉会有个坑，就是答案为2时，除了18-小的和的两个最小数  以及  大的和的2个最大数 还有 一个打的和最大+9-一个小的和最小。

思路就是判1时 ，当然为 大的和的最大值 或者 小的和的9-最小  为活动范围。这写起来有点麻烦，看代码把

#include 
#include 
#include 
using namespace std;
int a[220];
int main()
{

    int A[3],B[3],a[3],b[3];
    for(int i=0;i<6;i++){
        char x;
        scanf("%c",&x);
        if(i<3){ A[i]=x-'0'; }
        else B[i-3]=x-'0';
    }
    int sum1,sum2;
    int s1=A[0]+A[1]+A[2];
    int s2=B[0]+B[1]+B[2];
    if(s1>s2){
        b[0]=A[0];
        b[1]=A[1];
        b[2]=A[2];
        a[0]=B[0];
        a[1]=B[1];
        a[2]=B[2];
        sum2=s1;
        sum1=s2;
    }
    else {
        b[0]=B[0];
        b[1]=B[1];
        b[2]=B[2];
        a[0]=A[0];
        a[1]=A[1];
        a[2]=A[2];
        sum2=s2;
        sum1=s1;
    }


    sort(a,a+3);
    sort(b,b+3);


    int tmp=abs(sum1-sum2);
    if(tmp==0) printf("0");
    else if(tmp<=b[2]||tmp<=(9-a[0])) printf("1");
    else if(tmp<=b[2]+b[1]||tmp<=(18-a[0]-a[1])||tmp<=b[2]+9-a[0]) printf("2");
    else printf("3");
    return 0;
}