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;
string s, t;
int n, m, i, j, k, u, v, a, b, dp[3001][3001];
int presums[26][3001], presumt[26][3001];
int aa, bb;
int main() {
  cin >> aa;
  for (bb = 0; bb < aa; bb++) {
    cin >> n >> s >> t;
    string s1 = s, t1 = t;
    sort(s1.begin(), s1.end());
    sort(t1.begin(), t1.end());
    if (s1 != t1) {
      cout << -1 << endl;
      continue;
    }
    for (i = 1; i <= n; i++) {
      for (j = 0; j < 26; j++) presums[j][i] = presums[j][i - 1];
      presums[s[i - 1] - 97][i]++;
    }
    for (i = 1; i <= n; i++) {
      for (j = 0; j < 26; j++) presumt[j][i] = presumt[j][i - 1];
      presumt[t[i - 1] - 97][i]++;
    }
    for (i = n; i >= 0; i--)
      for (j = n; j >= 0; j--) {
        dp[i][j] = -1e9;
        if ((i == n) or (j == n)) {
          dp[i][j] = 0;
          continue;
        }
        int fail = 0;
        for (k = 0; k < 26; k++)
          if (presums[k][i] > presumt[k][j]) fail = 1;
        if ((s[i] == t[j]) and (!fail))
          dp[i][j] = max(dp[i][j], dp[i + 1][j + 1] + 1);
        dp[i][j] = max(dp[i][j], dp[i + 1][j]);
        dp[i][j] = max(dp[i][j], dp[i][j + 1]);
      }
    cout << n - dp[0][0] << endl;
  }
}
