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 solve(int N, string& S, string& T) {
  vector<int> cnt(26, 0);
  int min_j = -1;
  vector<int> dp(N + 1), nx(N + 1);
  for (int i = 0; i < N; i++) {
    --cnt[S[i] - 'a'];
    while (cnt[S[i] - 'a'] < 0) {
      ++min_j;
      if (min_j == N) {
        return -1;
      }
      ++cnt[T[min_j] - 'a'];
    }
    int cur_best = N - 1;
    for (int j = min_j; j <= N; j++) {
      cur_best = min(cur_best, dp[j] + 1);
      nx[j] = cur_best;
      if (j < N && T[j] == S[i]) {
        cur_best = dp[j];
      }
    }
    swap(dp, nx);
  }
  return dp[N];
}
int main() {
  ios::sync_with_stdio(false);
  int T;
  cin >> T;
  while (T--) {
    int N;
    cin >> N;
    string S, T;
    cin >> S >> T;
    int ans = solve(N, S, T);
    cout << ans << '\n';
  }
}
