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
def LCS(X, Y, m, n):

    maxLength = 0           # stores the max length of LCS
    endingIndex = m         # stores the ending index of LCS in `X`

    # `lookup[i][j]` stores the length of LCS of substring `X[0…i-1]` and `Y[0…j-1]`
    lookup = [[0 for x in range(n + 1)] for y in range(m + 1)]

    # fill the lookup table in a bottom-up manner
    for i in range(1, m + 1):
        for j in range(1, n + 1):

            # if the current character of `X` and `Y` matches
            if X[i - 1] == Y[j - 1]:
                lookup[i][j] = lookup[i - 1][j - 1] + 1

                # update the maximum length and ending index
                if lookup[i][j] > maxLength:
                    maxLength = lookup[i][j]
                    endingIndex = i

    # return longest common substring having length `maxLength`
    return X[endingIndex - maxLength: endingIndex]


for _ in range(int(input())):
    a=input()
    b=input()
    n=len(a)
    m=len(b)
    an=(LCS(a, b, n, m))
    val=0
    ans=len(an)
    
    val+=n-ans
    val+=m-ans
    print(val)