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;
void chmax(int& a, int b) { a = max(a, b); }
void solve() {
  int N;
  string S, T;
  cin >> N >> S >> T;
  vector<vector<int>> numS(N + 1, vector<int>(26)),
      numT(N + 1, vector<int>(26));
  for (int i = 0; i < N; i++) {
    numS[i + 1] = numS[i];
    numT[i + 1] = numT[i];
    numS[i + 1][S[i] - 'a']++;
    numT[i + 1][T[i] - 'a']++;
  }
  if (numS[N] != numT[N]) {
    cout << -1 << endl;
    return;
  }
  vector<vector<int>> dp(N + 1, vector<int>(N + 1, -1e9));
  dp[0][0] = 0;
  for (int i = 0; i <= N; i++)
    for (int j = 0; j <= N; j++) {
      if (i < N) chmax(dp[i + 1][j], dp[i][j]);
      if (j < N) chmax(dp[i][j + 1], dp[i][j]);
      if (i < N && j < N && i <= j && S[i] == T[j]) {
        bool ok = true;
        for (int a = 0; a < 26; a++)
          if (numS[i][a] > numT[j][a]) ok = false;
        if (ok) chmax(dp[i + 1][j + 1], dp[i][j] + 1);
      }
    }
  int ans = N - dp[N][N];
  cout << ans << endl;
}
int main() {
  int T;
  cin >> T;
  while (T--) solve();
  return 0;
}
