B. Delete from the Left

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本文介绍了一种优化的字符串匹配算法，通过比较两个字符串的左端字符并逐步删除不匹配的字符，来实现最少操作次数使两字符串相等的目标。算法特别适用于处理大量文本数据的情况。

You are given two strings $s$ and $t$ . In a single move, you can choose any of two strings and delete the first (that is, the leftmost) character. After a move, the length of the string decreases by $1$ . You can't choose a string if it is empty.

For example:

by applying a move to the string "where", the result is the string "here",
by applying a move to the string "a", the result is an empty string "".

You are required to make two given strings equal using the fewest number of moves. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the initial strings.

Write a program that finds the minimum number of moves to make two given strings $s$ and $t$ equal.

Input

The first line of the input contains $s$ . In the second line of the input contains $t$ . Both strings consist only of lowercase Latin letters. The number of letters in each string is between 1 and $2 \cdot 10^{5}$ , inclusive.

Output

Output the fewest number of moves required. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the given strings.

Examples
inputCopy
test
west
outputCopy
2
inputCopy
codeforces
yes
outputCopy
9
inputCopy
test
yes
outputCopy
7
inputCopy
b
ab
outputCopy
1

Note

In the first example, you should apply the move once to the first string and apply the move once to the second string. As a result, both strings will be equal to "est".

In the second example, the move should be applied to the string "codeforces" $8$ times. As a result, the string becomes "codeforces" $\to$ "es". The move should be applied to the string "yes" once. The result is the same string "yes" $\to$ "es".

In the third example, you can make the strings equal only by completely deleting them. That is, in the end, both strings will be empty.

In the fourth example, the first character of the second string should be deleted.

代码：

#include <iostream>
#include <cstring>
#include <cmath>

using namespace std;

int main()
{
    string a,b;
    cin>>a>>b;
    int lena=a.length();
    int lenb=b.length();
    int ans=lena+lenb;
    if(lena>=lenb){
        for(int i=lena-1;i>=0;--i)
        {
            if(a[i]==b[--lenb]){ans-=2;}
            else break;
        }
    }
    else{
        for(int i=lenb-1;i>=0;--i)
        {
            if(b[i]==a[--lena]){ans-=2;}
            else break;
        }
    }
    cout<<ans<<endl;
    return 0;
}