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

import java.io.*;

public class GFG {
 /*
    static int lcs(String X, String Y ,int i, int j, int count){

 
        if (i == 0 || j == 0)
        {
            return count;
        }
 
        if (X.charAt(i - 1)
            == Y.charAt(j - 1))
        {
            count = lcs(X,Y,i - 1, j - 1, count + 1);
        }
        count = Math.max(count,
                         Math.max(lcs(X,Y,i, j - 1, 0),
                                  lcs(X,Y,i - 1, j, 0)));
        return count;
    }*/
    
static int LCSubStr(char X[], char Y[],
                         int m, int n)
    {
        // Create a table to store
        // lengths of longest common
        // suffixes of substrings.
        // Note that LCSuff[i][j]
        // contains length of longest
        // common suffix of
        // X[0..i-1] and Y[0..j-1].
        // The first row and first
        // column entries have no
        // logical meaning, they are
        // used only for simplicity of program
        int LCStuff[][] = new int[m + 1][n + 1];
       
        // To store length of the longest
        // common substring
        int result = 0;
 
        // Following steps build
        // LCSuff[m+1][n+1] in bottom up fashion
        for (int i = 0; i <= m; i++)
        {
            for (int j = 0; j <= n; j++)
            {
                if (i == 0 || j == 0)
                    LCStuff[i][j] = 0;
                else if (X[i - 1] == Y[j - 1])
                {
                    LCStuff[i][j]
                        = LCStuff[i - 1][j - 1] + 1;
                    result = Integer.max(result,
                                         LCStuff[i][j]);
                }
                else
                    LCStuff[i][j] = 0;
            }
        }
        return result;
    }
 
public static void main(String[] args) throws IOException
    {   BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int t=Integer.parseInt(br.readLine());
        
        while(t-- !=0){
        String X=br.readLine();
        String Y=br.readLine();
        
        int n, m;
    
 
        n = X.length();
        m = Y.length();

        int ans=n+ m -(2*LCSubStr(X.toCharArray(),Y.toCharArray(),n,m));
        System.out.println(ans);
        
        }
    }
}
    
