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 d[2005][2005];
int n;
string a, b;
int za[2005][26], zb[2005][26];
bool valid() {
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 26; j++) {
      za[i + 1][j] = za[i][j];
      zb[i + 1][j] = zb[i][j];
    }
    za[i + 1][a[i] - 'a']++;
    zb[i + 1][b[i] - 'a']++;
  }
  for (int i = 0; i < 26; i++)
    if (za[n][i] != zb[n][i]) return false;
  return true;
}
void mx(int& x, int y) { x = max(x, y); }
bool excess(int i, int j) {
  for (int k = 0; k < 26; k++) {
    if (k == a[i] - 'a') {
      if (za[i][k] >= zb[j][k]) return false;
    } else {
      if (za[i][k] > zb[j][k]) return false;
    }
  }
  return true;
}
int main() {
  ios::sync_with_stdio(!cin.tie(0));
  int t;
  cin >> t;
  while (t--) {
    cin >> n >> a >> b;
    for (int i = 0; i <= n; i++)
      for (int j = 0; j <= n; j++) d[i][j] = -123123123;
    d[0][0] = 0;
    if (!valid()) {
      cout << "-1\n";
      continue;
    }
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        if (a[i] == b[j]) {
          mx(d[i + 1][j + 1], d[i][j] + 1);
        }
        mx(d[i][j + 1], d[i][j]);
        if (excess(i, j)) mx(d[i + 1][j], d[i][j]);
      }
    }
    int h = 0;
    for (int j = 0; j <= n; j++) mx(h, d[j][n]);
    cout << n - h << '\n';
  }
}
