/*==================================================*\
| 最短公共祖先（两个长字符串）
| The shortest common superstring of 2 strings S1 and S2 is
| a string S with|the minimum number of characters which
| contains both S1 and S2 as a sequence of consecutive
| characters. HDU 1841
\*==================================================*/
#include <iostream>
#include <string>
using namespace std;
const int N = 1000010;
char a[2][N]; int fail[N];
inline int max(int a, int b) 
{ return ( a > b ) ? a : b; }
int kmp(int &i, int &j, char* str, char* pat)
{
int k;
memset(fail, -1, sizeof(fail));
for (i = 1; pat[i]; ++i)
{
for (k=fail[i-1]; k>=0 && pat[i]!=pat[k+1];
k=fail[k]);
if (pat[k + 1] == pat[i]) fail[i] = k + 1;
}
i = j = 0;
while( str[i] && pat[j] )
{
if( pat[j] == str[i] )
++i, ++j;
else if(j == 0)
++i;//第一个字符匹配失败，从 str下个字符开始
else
j = fail[j-1]+1;
}
if( pat[j] )
return -1;
else
return i-j;
}
int main(void)
{
int T;
scanf("%d\n", &T);
while( T-- )
{
int i, j, l1=0, l2=0;
gets(a[0]); gets(a[1]);
int len1 = strlen(a[0]), len2 = strlen(a[1]), val;
val = kmp(i, j, a[1], a[0]); // a[1]在前
if( val != -1 )
l1 = len1;
else
{
//printf("i:%d, j:%d\n", i, j);
if( i == len2 && j-1 >= 0 && a[1][len2-1] == a[0][j-1] )
l1 = j;
}
val = kmp(i, j, a[0], a[1]); // a[0]在前
if( val != -1 ) l2 = len2;
else
{
//printf("i:%d, j:%d\n", i, j);
if( i == len1 && j-1 >= 0 && a[0][len1-1] == a[1][j-1] )
l2 = j;
}
// printf("l1:%d, l2:%d\n", l1, l2);
printf("%d\n", len1+len2-max(l1, l2));
}
return 0;
}