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.
INF = float('inf')


def solve():
    N = int(input())
    A = input().strip()
    B = input().strip()

    Acount = [[0] * 26 for _ in range(N + 1)]
    Bcount = [[0] * 26 for _ in range(N + 1)]
    for i in range(N - 1, -1, -1):
        Acount[i][ord(A[i]) - ord('a')] += 1
        Bcount[i][ord(B[i]) - ord('a')] += 1
        for j in range(26):
            Acount[i][j] += Acount[i + 1][j]
            Bcount[i][j] += Bcount[i + 1][j]

    for i in range(26):
        if Acount[0][i] != Bcount[0][i]:
            return -1

    dp = [[INF] * (N + 1) for _ in range(N + 1)]
    for j in range(N + 1):
        dp[0][j] = 0
    
    for i in range(1, N + 1):
        s = ord(A[i - 1]) - ord('a')
        for j in range(i, N + 1):
            dp[i][j] = min(dp[i][j], dp[i - 1][j] + 1)

            t = ord(B[j - 1]) - ord('a')
            if Acount[i][t] > Bcount[j][t]:
                dp[i][j] = min(dp[i][j], dp[i][j - 1])
            if s == t:
                dp[i][j] = min(dp[i][j], dp[i - 1][j - 1])

    # for row in dp:
    #     print(row)

    return dp[N][N]


if __name__ == '__main__':
    T = int(input())
    for _ in range(T):
        print(solve())