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 tc, n, dp[2002][2002], vis[2002][2002], tcc, f1[2002][300], f2[2002][300];
string s, t;
int solve(int i, int j) {
  if (j + i == 0) return 0;
  int &ans = dp[i][j];
  if (vis[i][j] == tc) return ans;
  vis[i][j] = tc;
  ans = 1e5;
  if (j && f1[i + 1][t[j - 1]] > f2[j + 1][t[j - 1]]) ans = solve(i, j - 1);
  if (j && i && s[i - 1] == t[j - 1]) ans = min(ans, solve(i - 1, j - 1));
  if (i) ans = min(ans, solve(i - 1, j) + 1);
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> tcc;
  while (tcc--) {
    cin >> n >> s >> t;
    tc++;
    for (int j = 'a'; j <= 'z'; j++) f1[n + 1][j] = f2[n + 1][j] = 0;
    for (int i = n; i >= 1; i--) {
      for (int j = 'a'; j <= 'z'; j++) {
        f1[i][j] = f1[i + 1][j] + (s[i - 1] == j);
        f2[i][j] = f2[i + 1][j] + (t[i - 1] == j);
      }
    }
    int ans = solve(n, n);
    if (ans >= 1e5) ans = -1;
    cout << ans << "\n";
  }
}
