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 read() {
  int xx = 0, flagg = 1;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') {
    flagg = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    xx = xx * 10 + ch - '0';
    ch = getchar();
  }
  return xx * flagg;
}
void pus(int xx, int flagg) {
  if (xx < 0) {
    putchar('-');
    xx = -xx;
  }
  if (xx >= 10) pus(xx / 10, 0);
  putchar(xx % 10 + '0');
  if (flagg == 1) putchar(' ');
  if (flagg == 2) putchar('\n');
  return;
}
int t, n, i, j, k, book[2005], l[2005], f[2005][2005], ans;
char s[2005], c[2005];
int main() {
  t = read();
  while (t--) {
    n = read();
    scanf("%s%s", s + 1, c + 1);
    for (i = 0; i <= 25; i++) book[i] = 0;
    for (i = 1; i <= n; i++) {
      book[s[i] - 'a']++;
      book[c[i] - 'a']--;
    }
    for (i = 0; i <= 25; i++)
      if (book[i] != 0) {
        pus(-1, 2);
        break;
      }
    if (i == 26) {
      i = n + 1;
      for (j = n; j >= 1; j--) {
        book[c[j] - 'a']--;
        while (book[c[j] - 'a'] < 0 && i != 0) {
          i--;
          book[s[i] - 'a']++;
        }
        l[j] = i;
      }
      for (i = 1; i <= n; i++)
        for (j = 1; j <= n; j++) {
          f[i][j] = max(f[i - 1][j], f[i][j - 1]);
          if (s[i] == c[j] && i <= l[j])
            f[i][j] = max(f[i][j], f[i - 1][j - 1] + 1);
        }
      pus(n - f[n][n], 2);
    }
  }
  return 0;
}
