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 N = 4e5 + 10;
int suffixa[2020][26];
int suffixb[2020][26];
int f[2010][2010];
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    for (int i = 0; i <= n; i++) {
      for (int j = 0; j < 26; j++) {
        suffixa[i][j] = 0;
        suffixb[i][j] = 0;
      }
    }
    for (int i = 0; i <= n; i++) {
      for (int j = 0; j <= n; j++) f[i][j] = 0;
    }
    int i;
    string s, t;
    cin >> s >> t;
    s = " " + s;
    t = " " + t;
    string tmp1 = s;
    string tmp2 = t;
    sort(tmp1.begin(), tmp1.end());
    sort(tmp2.begin(), tmp2.end());
    if (tmp1 != tmp2) {
      cout << -1 << endl;
      continue;
    }
    for (i = (int)s.size() - 1; i > 0; i--) {
      for (int j = 0; j < 26; j++) {
        suffixa[i][j] = suffixa[i + 1][j];
        suffixb[i][j] = suffixb[i + 1][j];
      }
      suffixa[i][s[i] - 'a']++;
      suffixb[i][t[i] - 'a']++;
    }
    for (i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
        f[i][j] = max(f[i][j], max(f[i - 1][j], f[i][j - 1]));
        if (s[i] == t[j]) {
          int jj;
          int sign = 0;
          for (jj = 0; jj < 26; jj++) {
            if (suffixa[i][jj] < suffixb[j][jj]) {
              sign = 1;
              break;
            }
          }
          if (!sign) f[i][j] = max(f[i][j], f[i - 1][j - 1] + 1);
        }
      }
    }
    cout << n - f[n][n] << endl;
  }
}
