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 int long long
#define pb push_back
#define mp make_pair
#define endl "\n"
#define trap_time() std::chrono::high_resolution_clock::now()
#define trap_time_diff(a,b) 1e-9 * std::chrono::duration_cast<std::chrono::nanoseconds>(a - b).count()
#define fr(i,s,k,b) for(int i=s;i<=k;i+=b)
#define fast() ios::sync_with_stdio(0);cin.tie(0);
#define mod 1000000007
using namespace std;
int32_t main()
{
    fast();
    int t=1;
    cin>>t;
    while(t--)
    {
        string a,b;
        cin>>a>>b;
        string mx,mn;
        if(a.size()>b.size())
        {
            mx=a;
            mn=b;
        }
        else
        {
            mx=b;
            mn=a;
        }
        int cntmx=0;
        for(int i=mn.size()-1;i>=0;i--)
        {
            for(int j=0;j<=i;j++)
            {
                string xx="";
                for(int k=j;k<=i;k++)
                    xx+=mn[k];
                for(int cv=0;cv<mx.size()-xx.size()+1;cv++)
                {
                    string zx="";
                    for(int cc=cv;cc<cv+xx.size();cc++)
                        zx+=mx[cc];
                    if(zx==xx)
                    {
                        if(xx.size()>cntmx)
                        {
                            cntmx=xx.size();
                            break;
                        }
                    }
                }
            }
        }
        int ans=a.size()+b.size()-2*cntmx;
        cout<<ans<<endl;
    }
}
