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 long long MX = 2011;
long long precal[2][MX][26];
bool used[MX][MX];
long long memo[MX][MX];
string a, b;
bool can(string p, string q) {
  sort(p.begin(), p.end());
  sort(q.begin(), q.end());
  return p == q;
}
long long dp(long long l, long long r) {
  if (l == 0 && r == 0) return 0;
  if (used[l][r]) return memo[l][r];
  used[l][r] = true;
  long long ans = 1e18;
  if (l > 0 && r > 0 && a[l - 1] == b[r - 1]) ans = min(ans, dp(l - 1, r - 1));
  if (r > 0 && precal[1][r][b[r - 1] - 'a'] < precal[0][l][b[r - 1] - 'a'])
    ans = min(ans, dp(l, r - 1));
  if (l > 0) ans = min(ans, dp(l - 1, r) + 1);
  memo[l][r] = ans;
  return ans;
}
void solve() {
  long long n;
  cin >> n;
  cin >> a >> b;
  if (!can(a, b)) {
    cout << -1 << "\n";
    return;
  }
  for (long long i = 0; i < 26; i++) {
    for (long long j = 0; j <= n; j++) {
      precal[0][j][i] = 0;
      precal[1][j][i] = 0;
    }
  }
  for (long long i = 0; i <= n; i++) {
    for (long long j = 0; j <= n; j++) {
      memo[i][j] = used[i][j] = false;
    }
  }
  for (long long i = 0; i < n; i++) {
    precal[0][i][a[i] - 'a'] = 1;
    precal[1][i][b[i] - 'a'] = 1;
  }
  for (long long j = 0; j < 26; j++) {
    for (long long i = n - 1; i >= 0; i--) {
      precal[0][i][j] = precal[0][i][j] + precal[0][i + 1][j];
      precal[1][i][j] = precal[1][i][j] + precal[1][i + 1][j];
    }
  }
  cout << dp(n, n) << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout.precision(10);
  cout << fixed;
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}
