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;
const int MX = 2e3 + 5;
string s, t;
int dp[MX][MX], cnt_s[MX][26], cnt_t[MX][26];
int func(int ps, int pt) {
  if (ps < 0) return 0;
  int &r = dp[ps][pt];
  if (r != -1) return r;
  r = 1 + func(ps - 1, pt);
  if (s[ps] == t[pt]) r = min(r, func(ps - 1, pt - 1));
  if (cnt_s[ps][t[pt] - 'a'] < cnt_t[pt][t[pt] - 'a'])
    r = min(r, func(ps, pt - 1));
  return r;
}
int main() {
  int i, j, k;
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    cin >> s >> t;
    for (i = 0; i < n; ++i)
      for (j = 0; j < n; ++j) dp[i][j] = -1;
    int flag = 0;
    for (i = 0; i < n; ++i) {
      for (j = 0; j < 26; ++j) {
        if (i)
          cnt_s[i][j] = cnt_s[i - 1][j], cnt_t[i][j] = cnt_t[i - 1][j];
        else
          cnt_s[i][j] = 0, cnt_t[i][j] = 0;
        cnt_s[i][j] += (j == s[i] - 'a');
        cnt_t[i][j] += (j == t[i] - 'a');
        if (i == n - 1 && cnt_s[i][j] != cnt_t[i][j]) flag = 1;
      }
    }
    if (flag)
      cout << "-1" << endl;
    else
      cout << func(n - 1, n - 1) << endl;
  }
}
