Lucky Conversion（找规律）

Description

Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

Petya has two strings a and b of the same length n. The strings consist only of lucky digits. Petya can perform operations of two types:

• replace any one digit from string a by its opposite (i.e., replace 4 by 7 and 7 by 4);
• swap any pair of digits in string a.

Petya is interested in the minimum number of operations that are needed to make string a equal to string b. Help him with the task.

Input

The first and the second line contains strings a and b, correspondingly. Strings a and b have equal lengths and contain only lucky digits. The strings are not empty, their length does not exceed 10⁵.

Output

Print on the single line the single number — the minimum number of operations needed to convert string a into string b.

Sample Input

Input

47
74

Output

1

Input

774
744

Output

1

Input

777
444

Output

3

题目意思：给你两个字符串a,b，a,b都是由4和7组成的。a串中的4和7可以通过交换位置或者替换（4替换成7,7替换成4），
问最少多少步就能得到b串。

解题思路：这算是一道找规律的题了，其实可以这样做，统计a串和b串不同的字符在b串中表现是4还是7，记录一下个数，
4的个数和7的个数可以相互抵消来表示交换，剩下不能抵消的可以用来替换。抵消的次数加上替换的次数就是需要的步数。

上代码：

#include<stdio.h>
#include<string.h>
int main()
{
    char a[],b[];
    int k,i,count_7=,count_4=,count=;
    memset(a,,sizeof(a));
    memset(b,,sizeof(b));
    gets(a);
    gets(b);
    k=strlen(a);
    for(i=; i<k; i++)
    {
        if(a[i]==b[i])
            continue;
        else
        {
            if(b[i]=='')
                count_7++;
            else
                count_4++;
        }
    }
    if(count_7>=count_4)
        count=count_4+count_7-count_4;
    else if(count_7<count_4)
        count=count_7+count_4-count_7;
    printf("%d\n",count);
    return ;
}