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;
const int MX = 2005;
int n, mem[MX][MX], as[MX][256], at[MX][256];
string s, t, _s, _t;
int dp(int i, int j) {
  if (i > j) return 0;
  if (i == n || j == n) return 0;
  int &res = mem[i][j];
  if (res != -1) return res;
  res = 0;
  if (s[i] == t[j]) res = max(res, dp(i + 1, j + 1) + 1);
  if (as[i + 1][s[i]] >= at[j][s[i]]) res = max(res, dp(i + 1, j));
  if (as[i + 1][t[j]] >= at[j][t[j]]) res = max(res, dp(i, j + 1));
  return res;
}
void solve() {
  cin >> n >> s >> t;
  _s = s;
  _t = t;
  sort(_s.begin(), _s.end());
  sort(_t.begin(), _t.end());
  if (_s != _t) {
    cout << -1 << endl;
    return;
  }
  memset(as[n], 0, sizeof(as[n]));
  for (int i = n - 1; i >= 0; i--) {
    for (int j = 'a'; j <= 'z'; j++) as[i][j] = as[i + 1][j];
    as[i][s[i]]++;
  }
  memset(at[n], 0, sizeof(at[n]));
  for (int i = n - 1; i >= 0; i--) {
    for (int j = 'a'; j <= 'z'; j++) at[i][j] = at[i + 1][j];
    at[i][t[i]]++;
  }
  for (int i = 0; i < (n + 1); i++)
    for (int j = 0; j < (n + 1); j++) mem[i][j] = -1;
  cout << n - dp(0, 0) << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}
