Delete from the Left CodeForces - 1005B （水题）

B. Delete from the Left

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

input

Copy

test
west

output

Copy

2

input

Copy

codeforces
yes

output

Copy

9

input

Copy

test
yes

output

Copy

7

input

Copy

b
ab

output

Copy

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.

题意：给出两个字符串，你可以将这两个字符串的首字母删除（可多次操作），问最少删除多少个字母，可以使得这两个字符串相同（如果两个是两个空串，也认为他们相同）

思路：由于只能删除首字母，要使两个字符串相同，那么实际上最后的字符串就是从这两个字符串的尾部开始找最长公共子串。

#include "iostream"
#include "string"
using namespace std;
int main()
{
    string s,t;
    int cnt=0;
    cin>>s>>t;
    for(int i=s.size()-1,j=t.size()-1;i>=0&&j>=0;i--,j--){
        if(s[i]==t[j]) cnt++;
        else break;
    }
    cout<<s.size()-cnt+t.size()-cnt<<endl;
}