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;
typedef long long ll;
ll power(ll a, ll b){
    ll res=1;
    while(b){
        if(b&1) res=(res*a);
        a=(a*a);
        b>>=1;
    }
    return res;
}

int main()
{
	ll t;
	cin>>t;
	while(t--)
	{
		ll i,j,p,k;
		string a,b;
		cin>>a>>b;
		ll n=a.size();
		ll m=b.size();
		ll cnt=0;
		ll ans=0;
		for(i=n;i>0;i--)
		{
			for(j=0;j<=m-i;j++)
			{
				for(k=0;k<=n-i;k++)
				{
					for(p=j;p<j+i;p++)
					{
					if(a[p-j+k]==b[p])
					cnt++;
					}
					if(cnt==i)
					{
					ans=i;
					}
					cnt=0;
				}
			}
			if(ans>0)
			break;
		}
		cout<<n+m-2*ans<<endl;
	}
}
