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 read_int():
    return int(input())


def read_ints():
    return map(int, input().split(' '))


t = read_int()
for case_num in range(t):
    n = read_int()
    cnt = [0 for i in range(26)]
    ps = [[0 for j in range(n + 1)] for i in range(26)]
    pt = [[0 for j in range(n + 1)] for i in range(26)]
    s = input()
    t = input()
    for i in range(n):
        ch = ord(s[i]) - ord('a')
        cnt[ch] += 1
        for j in range(26):
            ps[j][i + 1] = ps[j][i] + (1 if ch == j else 0)
    for i in range(n):
        ch = ord(t[i]) - ord('a')
        cnt[ch] -= 1
        for j in range(26):
            pt[j][i + 1] = pt[j][i] + (1 if ch == j else 0)
    ok = True
    for i in cnt:
        if i != 0:
            ok = False
            break
    if not ok:
        print(-1)
    else:
        r = n
        while r >= 1 and s[r - 1] == t[r - 1]:
            r -= 1
        inf = int(1e8)
        dp = [[0 if i == 0 else inf for j in range(
            r + 1)] for i in range(r + 1)]
        for i in range(1, r + 1):
            for j in range(i, r + 1):
                if s[i - 1] == t[j - 1]:
                    dp[i][j] = min(dp[i][j], dp[i - 1][j - 1])
                ch = ord(t[j - 1]) - ord('a')
                if ps[ch][r] - ps[ch][i] > pt[ch][r] - pt[ch][j]:
                    dp[i][j] = min(dp[i][j], dp[i][j - 1])
                dp[i][j] = min(dp[i][j], dp[i - 1][j] + 1)
        print(dp[r][r])
