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>
const int MAXN = 2000 + 5;
char s[MAXN], t[MAXN];
int n;
int f[MAXN][MAXN];
int p1[MAXN][26], p2[MAXN][26];
inline void Solve() {
  scanf("%d", &n);
  scanf("%s%s", s + 1, t + 1);
  std::reverse(s + 1, s + n + 1);
  std::reverse(t + 1, t + n + 1);
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= 25; ++j)
      p1[i][j] = p1[i - 1][j], p2[i][j] = p2[i - 1][j];
    p1[i][s[i] - 'a']++;
    p2[i][t[i] - 'a']++;
  }
  for (int i = 0; i <= 25; ++i) {
    if (p1[n][i] != p2[n][i]) {
      puts("-1");
      return;
    }
  }
  for (int i = 0; i <= n + 1; ++i)
    for (int j = 0; j <= n + 1; ++j) f[i][j] = 1e9;
  for (int i = 0; i <= n; ++i) f[i][0] = i;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= i; ++j) {
      bool flag = true;
      for (int k = 0; k <= 25; ++k) flag &= p1[i][k] >= p2[j][k];
      if (!flag) continue;
      if (s[i] == t[j]) f[i][j] = std::min(f[i][j], f[i - 1][j - 1]);
      f[i][j] = std::min(f[i][j], f[i - 1][j] + 1);
      f[i][j] = std::min(f[i][j], f[i][j - 1]);
    }
  }
  printf("%d\n", f[n][n]);
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) Solve();
  return 0;
}
