You are given the strings a and b, consisting of lowercase Latin letters. You can do any number of the following operations in any order: 

  * if |a| > 0 (the length of the string a is greater than zero), delete the first character of the string a, that is, replace a with a_2 a_3 … a_n; 
  * if |a| > 0, delete the last character of the string a, that is, replace a with a_1 a_2 … a_{n-1}; 
  * if |b| > 0 (the length of the string b is greater than zero), delete the first character of the string b, that is, replace b with b_2 b_3 … b_n; 
  * if |b| > 0, delete the last character of the string b, that is, replace b with b_1 b_2 … b_{n-1}. 



Note that after each of the operations, the string a or b may become empty.

For example, if a="hello" and b="icpc", then you can apply the following sequence of operations: 

  * delete the first character of the string a ⇒ a="ello" and b="icpc"; 
  * delete the first character of the string b ⇒ a="ello" and b="cpc"; 
  * delete the first character of the string b ⇒ a="ello" and b="pc"; 
  * delete the last character of the string a ⇒ a="ell" and b="pc"; 
  * delete the last character of the string b ⇒ a="ell" and b="p". 



For the given strings a and b, find the minimum number of operations for which you can make the strings a and b equal. Note that empty strings are also equal.

Input

The first line contains a single integer t (1 ≤ t ≤ 100). Then t test cases follow.

The first line of each test case contains the string a (1 ≤ |a| ≤ 20), consisting of lowercase Latin letters.

The second line of each test case contains the string b (1 ≤ |b| ≤ 20), consisting of lowercase Latin letters.

Output

For each test case, output the minimum number of operations that can make the strings a and b equal.

Example

Input


5
a
a
abcd
bc
hello
codeforces
hello
helo
dhjakjsnasjhfksafasd
adjsnasjhfksvdafdser


Output


0
2
13
3
20
#include <bits/stdc++.h>
using namespace std;
 
int main()
{
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);	
#endif 
  	// CODE FROM HERE
  	int tc;
  	cin>> tc;
  	for(int kd=0; kd<tc; kd++)
  	{
  		string x ,y;
		cin >> x >> y;
		int n=x.length();
    	int m=y.length();
    	int lcs[n+1][m+1];
    	int result=0;
    	for(int i=0;i<n+1;i++)
    		{
        		for(int j=0;j<m+1;j++)
        			{
            			if(i==0||j==0)
            				{
                				lcs[i][j]=0;
            				}
            			else if(x[i-1]==y[j-1])
            				{
	            				lcs[i][j]=1+lcs[i-1][j-1];
	            				result =max(result,lcs[i][j]);
            				}
            			else 
            				lcs[i][j]=0;
        			}
    		}
    	cout << (n+m)-(2*result) << endl; 	
  	}
  	
	
} 


	
	
