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;
struct fast_ios {
  fast_ios() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(20);
  };
} fast_ios_;
template <typename T>
bool chmax(T &m, const T q) {
  if (m < q) {
    m = q;
    return true;
  } else
    return false;
}
template <typename T>
bool chmin(T &m, const T q) {
  if (m > q) {
    m = q;
    return true;
  } else
    return false;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &vec) {
  for (auto &v : vec) is >> v;
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &vec) {
  os << "[";
  for (auto v : vec) os << v << ",";
  os << "]";
  return os;
}
constexpr int INF = 1e9;
int solve() {
  int N;
  string S, T;
  cin >> N >> S >> T;
  string ss = S, tt = T;
  sort((ss).begin(), (ss).end());
  sort((tt).begin(), (tt).end());
  if (ss != tt) return -1;
  vector<int> pend(26);
  vector<vector<int>> scnt(26, vector<int>(N + 1));
  vector<vector<int>> tcnt(26, vector<int>(N + 1));
  for (int i = (0), i_end_ = (N); i < i_end_; i++) {
    for (int d = (0), d_end_ = (26); d < d_end_; d++)
      scnt[d][i + 1] = scnt[d][i];
    scnt[S[i] - 'a'][i + 1]++;
    for (int d = (0), d_end_ = (26); d < d_end_; d++)
      tcnt[d][i + 1] = tcnt[d][i];
    tcnt[T[i] - 'a'][i + 1]++;
  }
  vector<vector<int>> dp(N + 2, vector<int>(N + 2, -INF));
  dp[0][0] = 0;
  for (int i = (0), i_end_ = (N + 1); i < i_end_; i++)
    for (int j = (0), j_end_ = (N + 1); j < j_end_; j++)
      if (dp[i][j] > -INF) {
        if (i < N and j < N and S[i] == T[j])
          chmax(dp[i + 1][j + 1], dp[i][j] + 1);
        if (i < N and scnt[S[i] - 'a'][i + 1] <= tcnt[S[i] - 'a'][j])
          chmax(dp[i + 1][j], dp[i][j]);
        chmax(dp[i][j + 1], dp[i][j]);
      }
  return N - dp[N][N];
}
int main() {
  int T;
  cin >> T;
  while (T--) cout << solve() << '\n';
}
