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 ini() {
  int x;
  scanf("%d", &x);
  return x;
}
long long inl() {
  long long x;
  scanf("%lld", &x);
  return x;
}
void file_mamager() {}
char ss[20005], tt[20005];
int scn[30][20005], tcn[30][20005];
int main() {
  int t = ini();
  while (t--) {
    int n = ini();
    for (int i = 1; i <= n; i++) {
      cin >> ss[i];
      for (int j = 0; j < 26; j++) {
        if (j == (ss[i] - 'a'))
          scn[j][i] = 1;
        else
          scn[j][i] = 0;
        scn[j][i] += scn[j][i - 1];
      }
    }
    for (int i = 1; i <= n; i++) {
      cin >> tt[i];
      for (int j = 0; j < 26; j++) {
        if (j == (tt[i] - 'a'))
          tcn[j][i] = 1;
        else
          tcn[j][i] = 0;
        tcn[j][i] += tcn[j][i - 1];
      }
    }
    bool possible = true;
    for (int j = 0; j < 26; j++)
      if (scn[j][n] != tcn[j][n]) possible = false;
    if (!possible) {
      cout << "-1\n";
      continue;
    }
    vector<vector<int> > dp(n + 5, vector<int>(n + 5));
    for (int i = 0; i <= n; i++) {
      for (int j = 0; j <= n; j++) dp[i][j] = (i ? 1e8 : 0);
    }
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
        if (ss[i] == tt[j]) dp[i][j] = dp[i - 1][j - 1];
        if (scn[ss[i] - 'a'][i - 1] < tcn[ss[i] - 'a'][j])
          dp[i][j] = min(dp[i][j], dp[i - 1][j] + 1);
        if (i <= j - 1) dp[i][j] = min(dp[i][j], dp[i][j - 1]);
      }
    }
    cout << dp[n][n] << "\n";
  }
  return 0;
}
