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.
#include <bits/stdc++.h>
using namespace std;
int n, l, dp[2010][2010], nums[2010][226], numt[2010][226];
string s, t;
void prep() {
  reverse(s.begin(), s.end());
  s = "0" + s;
  reverse(t.begin(), t.end());
  t = "0" + t;
  for (int i = 0; i <= l; i++)
    for (int j = 0; j <= l; j++) dp[i][j] = 1e5;
  for (int i = 0; i <= l; i++) dp[i][0] = i;
}
int count_and_judge() {
  for (int i = 1; i <= l; i++)
    for (int j = 'a'; j <= 'z'; j++) {
      nums[i][j] = nums[i - 1][j] + (s[i] == j);
      numt[i][j] = numt[i - 1][j] + (t[i] == j);
    }
  for (int i = 'a'; i <= 'z'; i++)
    if (nums[l][i] != numt[l][i]) return 0;
  return 1;
}
int work() {
  if (count_and_judge() == 0) return -1;
  for (int i = 1; i <= l; i++)
    for (int j = 1; j <= i; j++) {
      int flag = 0;
      for (int k = 'a'; k <= 'z'; k++)
        if (nums[i][k] < numt[j][k]) flag = 1;
      if (flag) continue;
      if (dp[i - 1][j] < 1e5) dp[i][j] = min(dp[i][j], dp[i - 1][j] + 1);
      if (dp[i][j - 1] < 1e5) dp[i][j] = min(dp[i][j], dp[i][j - 1]);
      if (s[i] == t[j] && dp[i - 1][j - 1] < 1e5)
        dp[i][j] = min(dp[i][j], dp[i - 1][j - 1]);
    }
  return dp[l][l];
}
int main() {
  cin >> n;
  while (n--) {
    cin >> l >> s >> t;
    prep();
    cout << work() << endl;
  }
  return 0;
}
