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>
#define ll long long
#define pb push_back 
//#define for(it , a) for(map<int,int>::iterator it=a.begin();it!=a.end();it++)
//#define fro(i, a, b)      for(__typeof(b) i=(a); i<(b); i++)
//upper_bound(a, a+n, k)-a
//    *min_element(iterator start, iterator end);
using namespace std;
void solve()
{
	string a, b;
	cin >> a >> b;
	int n = a.size(), m=b.size();
	if ( a == b)
	{
		cout<<"0"<<endl;
		return;
	}
	int ans = 0;
	for(int l=1; l<=min(n, m); l++)
	{
		for(int i=0; i+l<=n; i++)
		{
			for(int j=0; j+l<=m; j++)
			{
				if(a.substr(i, l) == b.substr(j, l))
				{
					ans=max(ans, l);
				}
			}
		}
	}
	cout<<n+m-2*ans<<endl;
}
int main()
{
	ll t=1;
            	cin>>t;
	while(t--)
	{
		solve();
	}
}