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>
void err() { std::cout << std::endl; }
template <typename T, typename... Args>
void err(T a, Args... args) {
  std::cout << a << ' ';
  err(args...);
}
template <template <typename...> class T, typename t, typename... A>
void err(const T<t> &arg, const A &...args) {
  for (auto &v : arg) std::cout << v << ' ';
  err(args...);
}
using namespace std;
const int N = 2000 + 5;
void run() {
  int n;
  cin >> n;
  string s, t;
  cin >> s >> t;
  string ss = s, tt = t;
  sort((ss).begin(), (ss).end()), sort((tt).begin(), (tt).end());
  if (ss != tt) {
    cout << -1 << '\n';
    return;
  }
  s = '#' + s;
  t = '#' + t;
  vector<vector<int>> dp(n + 1, vector<int>(n + 1));
  vector<vector<int>> suf_s(n + 2, vector<int>(26)),
      suf_t(n + 2, vector<int>(26));
  for (int i = n; i >= 1; i--) {
    for (int j = 0; j < 26; j++) {
      suf_s[i][j] = suf_s[i + 1][j];
      suf_t[i][j] = suf_t[i + 1][j];
    }
    ++suf_s[i][s[i] - 'a'];
    ++suf_t[i][t[i] - 'a'];
  }
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i) dp[i][j] = dp[i - 1][j];
      if (j) dp[i][j] = max(dp[i][j], dp[i][j - 1]);
      if (i && j && s[i] == t[j]) {
        bool ok = true;
        for (int k = 0; k < 26; k++) {
          if (suf_s[i][k] < suf_t[j][k]) ok = false;
        }
        if (ok) {
          dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + 1);
        }
      }
    }
  }
  cout << n - dp[n][n] << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(20);
  int T;
  cin >> T;
  while (T--) run();
  return 0;
}
