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 N, tc;
int memo[2005][2005];
int pS[30][2005];
int pT[30][2005];
int S[2005], T[2005];
void input() {
  char tp;
  for (int i = 1; i <= N; i++) {
    cin >> tp;
    S[i] = tp - 'a';
    for (int j = 0; j < 26; j++) {
      pS[j][i] = pS[j][i - 1];
    }
    pS[S[i]][i]++;
  }
  for (int i = 1; i <= N; i++) {
    cin >> tp;
    T[i] = tp - 'a';
    for (int j = 0; j < 26; j++) {
      pT[j][i] = pT[j][i - 1];
    }
    pT[T[i]][i]++;
  }
  S[N + 1] = 27, T[N + 1] = 27;
}
int dp(int p, int q) {
  if (q > N) return 0;
  if (memo[p][q] != -1) return memo[p][q];
  int res = N;
  if (S[p] == T[q]) res = min(res, dp(p + 1, q + 1));
  res = min(res, dp(p, q + 1) + 1);
  if (p < q && pS[S[p]][p - 1] < pT[S[p]][q - 1]) res = min(res, dp(p + 1, q));
  return memo[p][q] = res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> tc;
  while (tc--) {
    cin >> N;
    for (int i = 1; i <= N + 1; i++) {
      for (int j = 1; j <= N + 1; j++) {
        memo[i][j] = -1;
      }
    }
    char tp;
    input();
    bool ok = 1;
    for (int i = 0; i < 26; i++) {
      if (pS[i][N] != pT[i][N]) ok = 0;
    }
    if (!ok)
      cout << "-1\n";
    else
      cout << dp(1, 1) << "\n";
  }
}
