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 Len, cnts[2010][26], cntt[2010][26];
int f[2010][2010];
char s[2010], t[2010];
inline int read() {
  int x = 0, w = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    w |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return w ? -x : x;
}
int main() {
  int T = read();
  for (int lun = 1; lun <= T; lun++) {
    Len = read();
    scanf("%s", s + 1);
    scanf("%s", t + 1);
    memset(cnts[Len + 1], 0, 104);
    memset(cntt[Len + 1], 0, 104);
    for (int i = Len; i; i--) {
      memcpy(cnts[i], cnts[i + 1], 104);
      memcpy(cntt[i], cntt[i + 1], 104);
      cnts[i][s[i] - 'a']++;
      cntt[i][t[i] - 'a']++;
    }
    bool flag = 0;
    for (int i = 0; i < 26; i++)
      if (cnts[1][i] ^ cntt[1][i]) flag = 1;
    if (flag) {
      printf("-1\n");
      continue;
    }
    for (int i = 1; i <= Len; i++) memset(f[i], 0x3f, sizeof(int) * (Len + 5));
    memset(f[0], 0, sizeof(int) * (Len + 5));
    for (int i = 1; i <= Len; i++)
      for (int j = i; j <= Len; j++) {
        int temp = 0x3f3f3f3f;
        if (s[i] == t[j]) temp = f[i - 1][j - 1];
        if (cnts[i + 1][t[j] - 'a'] > cntt[j + 1][t[j] - 'a'])
          temp = min(temp, f[i][j - 1]);
        temp = min(f[i - 1][j] + 1, temp);
        f[i][j] = temp;
      }
    printf("%d\n", f[Len][Len]);
  }
}
