You are given two strings s and t, each of length n and consisting of lowercase Latin alphabets. You want to make s equal to t. 

You can perform the following operation on s any number of times to achieve it — 

  * Choose any substring of s and rotate it clockwise once, that is, if the selected substring is s[l,l+1...r], then it becomes s[r,l,l + 1 ... r - 1]. All the remaining characters of s stay in their position. 

For example, on rotating the substring [2,4] , string "abcde" becomes "adbce". 




A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.

Find the minimum number of operations required to convert s to t, or determine that it's impossible.

Input

The first line of the input contains a single integer t (1≤ t ≤ 2000) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer n (1≤ n ≤ 2000) — the length of the strings. 

The second and the third lines contain strings s and t respectively.

The sum of n over all the test cases does not exceed 2000.

Output

For each test case, output the minimum number of operations to convert s to t. If it is not possible to convert s to t, output -1 instead.

Example

Input


6
1
a
a
2
ab
ba
3
abc
cab
3
abc
cba
4
abab
baba
4
abcc
aabc


Output


0
1
1
2
1
-1

Note

For the 1-st test case, since s and t are equal, you don't need to apply any operation.

For the 2-nd test case, you only need to apply one operation on the entire string ab to convert it to ba.

For the 3-rd test case, you only need to apply one operation on the entire string abc to convert it to cab.

For the 4-th test case, you need to apply the operation twice: first on the entire string abc to convert it to cab and then on the substring of length 2 beginning at the second character to convert it to cba.

For the 5-th test case, you only need to apply one operation on the entire string abab to convert it to baba.

For the 6-th test case, it is not possible to convert string s to t.
def num(c):
    return ord(c) - 97

import sys
input = sys.stdin.readline

t = int(input())
for _ in range(t):
    n = int(input())
    s1 = input().strip()
    s2 = input().strip()

    char1 = [0] * 26
    char2 = [0] * 26
    for c in s1:
        char1[num(c)] += 1
    for c in s2:
        char2[num(c)] += 1
    if char1 != char2:
        print(-1)
        continue

    dp = [[(False, 0, 0) for j in range(n+1)] for i in range(n + 1)]
    dp[0][0] = [True, 0,[0]*26]

    def upd(a, b, val, sett):
        if not dp[a][b][0] or val > dp[a][b][1]:
            dp[a][b] = (True, val, sett)
    
    for i in range(n):
        for j in range(n):
            valid, val, tab = dp[i][j]
            if not valid:
                continue
            
            top = s1[i]
            bot = s2[j]

            if top == bot:
                #upd(i+1, j+1, val + 1, tab)
                if not dp[i + 1][j + 1][0] or val + 1 > dp[i + 1][j + 1][1]:
                    dp[i + 1][j + 1] = [True, val + 1, tab]

            if tab[num(top)] > 0:
                sett = tab[:]
                sett[num(top)] -= 1
                
                #upd(i+1, j, val, sett)
                if not dp[i + 1][j][0] or val > dp[i + 1][j][1]:
                    dp[i + 1][j] = [True, val, sett]

            sett = tab[:]
            sett[num(bot)] += 1
            #upd(i, j + 1, val, sett)
            if not dp[i][j + 1][0] or val > dp[i][j + 1][1]:
                dp[i][j + 1] = [True, val, sett]

            del dp[i][j][2]

    poss = [dp[i][n][1] for i in range(n + 1)]
        

    print(n - max(poss))
            
    
