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 <iostream>
#include <string>
#include <cstring>
using namespace std;
 
string LCS(string X, string Y, int m, int n)
{
    int maxlen = 0;         
    int endingIndex = m;    
 
    int lookup[m + 1][n + 1];
    memset(lookup, 0, sizeof(lookup));
    for (int i = 1; i <= m; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            if (X[i - 1] == Y[j - 1])
            {
                lookup[i][j] = lookup[i - 1][j - 1] + 1;
                if (lookup[i][j] > maxlen)
                {
                    maxlen = lookup[i][j];
                    endingIndex = i;
                }
            }
        }
    }
    return X.substr(endingIndex - maxlen, maxlen);
}
 
int main()
{
    int t;
    cin>>t;
    while(t--){
        string X,Y;
        cin>>X>>Y;
        int m = X.size(), n = Y.size();
        string ans=LCS(X, Y, m, n);
        int ok=ans.size();
        int c=(m-ok)+(n-ok);
        cout<<c<<"\n";
    }
    return 0;
}