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.
import os
### START FAST IO ###
os_input = os.read(0, int(1e7)).split()
os_input_pos = -1
answer_list = []
def read_s():
    global os_input_pos
    os_input_pos += 1
    return os_input[os_input_pos].decode()
def read_i():
    return int(read_s())
def write_s(v):
    answer_list.append(v)
def write_i(v):
    write_s(str(v))
def print_ans():
    os.write(1, "\n".join(answer_list).encode())
    os.write(1, "\n".encode())
#### END FAST IO ####

T = read_i()
while T:
  T -= 1
  n = read_i()
  s = read_s()
  t = read_s()
  if sorted(s) != sorted(t):
    write_i(-1)
    continue

  s_count = [[0 for i in range(n+1)] for j in range(26)]
  t_count = [[0 for i in range(n+1)] for j in range(26)]
  for i in range(n):
    for j in range(26):
      s_count[j][i] = s_count[j][i-1]
      t_count[j][i] = t_count[j][i-1]
    s_count[ord(s[i]) - ord('a')][i] += 1
    t_count[ord(t[i]) - ord('a')][i] += 1

  dp = [[0 for i in range(n+1)] for j in range(n+1)]
  for i in range(0, n):
    for j in range(i, n):
      dp[i][j] = dp[i-1][j] + 1
      if s[i] == t[j]:
        dp[i][j] = min(dp[i][j], dp[i-1][j-1])
      c = ord(t[j]) - ord('a')
      if s_count[c][i] < t_count[c][j]:
        dp[i][j] = min(dp[i][j], dp[i][j-1])
  write_i(dp[n-1][n-1])

print_ans()

