You are given two strings ss and tt. 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 11. 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 ssand tt equal.

Input

The first line of the input contains ss. In the second line of the input contains tt. Both strings consist only of lowercase Latin letters. The number of letters in each string is between 1 and 2⋅1052⋅105, 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

test
west

Output

2

Input

codeforces
yes

Output

9

Input

test
yes

Output

7

Input

b
ab

Output

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" 88times. As a result, the string becomes "codeforces" →→ "es". The move should be applied to the string "yes" once. The result is the same string "yes" →→ "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.

題意這個很好理解，但是要細心，千萬要細心，一定得細心，不能當讀題的好寶寶，我按照題目要求一步一步走就超時，超限，各種錯誤，最後比賽結束都沒寫出來，特意寫一下部落格來銘記。操作方法直接從字串的後面開始比，最後一位不相同直接輸出兩個字串的長度之和，否者相同一個記錄一個，然後用總長度之和減去相同的長度二倍就過了。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char a[200100],b[200100];
int main()
{
    while(~scanf("%s%s",a,b))
        {
            int l1=strlen(a);
            int l2=strlen(b);
            int tx=l1-1,ty=l2-1;
            int l=max(l1,l2);
            if(b[ty]!=a[tx])
                printf("%d\n",l1+l2);
            else
                {
                    int sum=0;
                    for(int i=0; i<l; i++)
                        {
                            if(b[ty--]==a[tx--])
                                sum++;
                            else 
                            break;
                        }
                    printf("%d\n",l1-sum+l2-sum);
                }
        }
    return 0;
}