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
#!/usr/bin/env python
# -*- coding: utf-8 -*-

"""Codeforces Round #710 (Div. 3)

Problem C. Double-ended Strings

:author:         Kitchen Tong
:mail:    kctong529@gmail.com

Please feel free to contact me if you have any question
regarding the implementation below.
"""

__version__ = '3.2'
__date__ = '2021-03-25'


import sys


def longest_common_substring(r, s):
    dp = [[0 for j in range(len(s)+1)] for i in range(2)]
    result = 0
    for i in range(len(r)):
        for j in range(len(s)):
            if r[i] == s[j]:
                dp[1][j+1] = dp[0][j] + 1
                result = max(dp[1][j+1], result)
            else:
                dp[1][j+1] = 0
        dp = [dp[1], dp[0]]
    return result

def solve(a, b):
    if a == b:
        return 0
    s = longest_common_substring(a, b)
    return len(a + b) - s - s


def main(argv=None):
    t = int(input())
    for _ in range(t):
        a = input()
        b = input()
        print(solve(a, b))
    return 0


if __name__ == "__main__":
    STATUS = main()
    sys.exit(STATUS)
