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 = 2005;
template <class T>
inline void read(T &x) {
  T f = 1;
  x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  x *= f;
}
char s[N], t[N];
int n, ans, st, S[N][26], T[N][26];
int f[N][N];
inline void solve() {
  read(n);
  ans = 0;
  scanf("%s %s", s + 1, t + 1);
  for (int i = (0); i <= (25); i++) S[n + 1][i] = T[n + 1][i] = 0;
  for (int i = (n); i >= (1); i--) {
    for (int j = (0); j <= (25); j++)
      S[i][j] = S[i + 1][j], T[i][j] = T[i + 1][j];
    S[i][s[i] - 'a']++;
    T[i][t[i] - 'a']++;
  }
  for (int i = (0); i <= (25); i++)
    if (S[1][i] != T[1][i]) return (void)(puts("-1"));
  for (int i = (1); i <= (n + 1); i++)
    for (int j = (1); j <= (n + 1); j++) f[i][j] = -1;
  f[n + 1][n + 1] = 0;
  for (int i = (n + 1); i >= (2); i--)
    for (int j = (n + 1); j >= (2); j--) {
      if (f[i][j] >= 0) {
        if (s[i - 1] == t[j - 1])
          f[i - 1][j - 1] = max(f[i - 1][j - 1], 1 + f[i][j]);
        else if (T[j - 1][t[j - 1] - 'a'] <= S[i - 1][t[j - 1] - 'a'])
          f[i][j - 1] = max(f[i][j - 1], f[i][j]);
        f[i - 1][j] = max(f[i - 1][j], f[i][j]);
      } else
        break;
    }
  ans = -1;
  for (int i = (1); i <= (n); i++) ans = max(ans, f[1][i]);
  printf("%d\n", n - ans);
}
int main() {
  int TEST;
  read(TEST);
  while (TEST--) solve();
}
