CodeForces - 1005B Delete from the Left

最新推荐文章于 2019-02-25 14:53:46 发布

原创最新推荐文章于 2019-02-25 14:53:46 发布 · 244 阅读

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本文介绍了一种算法，该算法通过删除字符的方式使两个字符串相等，并找到实现这一目标所需的最少步骤数。具体地，从两个字符串的左侧开始逐个删除字符，直到两个字符串完全相同或变为空字符串。

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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 ss and 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

Input

codeforces
yes

Output

Input

test
yes

Output

Input

b
ab

Output

#include<iostream>
#include<cmath>
#include<string.h>
#include<cstdio>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
int main(){
	string str1,str2;
	while(cin>>str1>>str2)
	{
	int cnt=0;
	reverse(str1.begin(),str1.end());
	reverse(str2.begin(),str2.end());
	for(int i=0;i<min(str1.size(),str2.size());i++)
	{
	if(str1[i]==str2[i])cnt++;
	else break;	
	}
	cout<<str1.size()+str2.size()-2*cnt;	
	} 
}