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;
class Debugger {
 public:
  template <typename T>
  void printVector(vector<T> const &vec, bool printSize = true) {
    if (printSize) {
      cout << vec.size() << endl;
    }
    for (auto &elem : vec) {
      cout << elem << " ";
    }
    cout << endl;
  }
  template <typename T>
  void printMatrix(vector<vector<T>> const &matrix) {
    cout << matrix.size() << " ";
    if (!matrix.empty()) {
      cout << matrix[0].size();
    }
    cout << endl;
    for (auto &vec : matrix) {
      printVector(vec, false);
    }
    cout << endl;
  }
};
const int INF = 1e9;
const int maxN = 100005;
class TaskF {
 private:
  Debugger debugger;

 public:
  void solveOne(int it) {
    int n;
    cin >> n;
    string s, t;
    cin >> s >> t;
    vector<vector<int>> acumS(n, vector<int>(26, 0)),
        acumT(n, vector<int>(26, 0));
    for (int i = 0; i < n; i++) {
      acumS[i][s[i] - 'a'] = 1;
      acumT[i][t[i] - 'a'] = 1;
      if (i > 0) {
        for (int alpha = 0; alpha < 26; alpha++) {
          acumS[i][alpha] += acumS[i - 1][alpha];
          acumT[i][alpha] += acumT[i - 1][alpha];
        }
      }
    }
    vector<vector<int>> dp(n + 1, vector<int>(n + 1, INF));
    dp[0][0] = 0;
    for (int i = 0; i <= n; i++) {
      for (int j = 0; j <= n; j++) {
        if (i > 0 and j > 0 and s[i - 1] == t[j - 1]) {
          dp[i][j] = min(dp[i][j], dp[i - 1][j - 1]);
        }
        if (i > 0 and j > 0) {
          dp[i][j] = min(dp[i][j], dp[i - 1][j] + 1);
        }
        if (j > 0) {
          int alpha = t[j - 1] - 'a';
          int sum1 = acumS[n - 1][alpha] - (i > 0 ? acumS[i - 1][alpha] : 0);
          int sum2 = acumT[n - 1][alpha] - acumT[j - 1][alpha];
          if (sum1 > sum2) {
            dp[i][j] = min(dp[i][j], dp[i][j - 1]);
          }
        }
      }
    }
    if (dp[n][n] >= INF) {
      cout << -1 << endl;
      return;
    }
    cout << dp[n][n] << endl;
  }
  void solve() {
    int tc = 1;
    cin >> tc;
    for (int it = 1; it <= tc; it++) {
      solveOne(it);
    }
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  TaskF solver;
  solver.solve();
  return 0;
}
