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;
int R = pow(10, 9) + 7;
template <typename T>
void print(vector<T>& v) {
  for (int i = 0; i < v.size(); i++) {
    cout << v[i] << ' ';
  }
  cout << endl;
}
template <typename T1, typename T2>
void print(vector<pair<T1, T2>>& v) {
  cout << "[";
  for (int i = 0; i < v.size(); i++) {
    cout << '(' << v[i].first << ',' << v[i].second << "),";
  }
  cout << "]" << endl;
}
template <typename T>
void print(vector<vector<T>>& v) {
  cout << "[";
  for (int i = 0; i < v.size(); i++) {
    print(v[i]);
  }
  cout << "]" << endl;
}
template <typename T>
void print(set<T>& v) {
  cout << "{";
  for (auto i : v) {
    cout << i << ' ';
  }
  cout << "}" << endl;
}
template <typename T1, typename T2>
void print(map<T1, T2>& v) {
  cout << "{";
  for (auto i : v) {
    cout << i.first << ':' << i.second << ',';
  }
  cout << "}" << endl;
}
template <typename T>
bool in(set<T>& indices, T x) {
  return indices.find(x) != indices.end();
}
template <typename T, typename T_>
bool in(map<T, T_>& indices, T x) {
  return indices.find(x) != indices.end();
}
int power(int x, int y) {
  int res = 1;
  x = x % R;
  if (x == 0) return 0;
  while (y > 0) {
    if (y % 2 == 1) res = (res * x) % R;
    y /= 2;
    x = (x * x) % R;
  }
  return res;
}
void TestCase() {
  int n;
  cin >> n;
  string s, t;
  cin >> s >> t;
  vector<vector<int>> count1(26, vector<int>(n + 1));
  vector<vector<int>> count2(26, vector<int>(n + 1));
  for (int i = 1; i <= n; i++) {
    int index = s[i - 1] - 'a';
    for (int j = 0; j < 26; j++) {
      count1[j][i] = count1[j][i - 1];
    }
    count1[index][i]++;
  }
  for (int i = 1; i <= n; i++) {
    int index = t[i - 1] - 'a';
    for (int j = 0; j < 26; j++) {
      count2[j][i] = count2[j][i - 1];
    }
    count2[index][i]++;
  }
  vector<vector<int>> dp(n + 1, vector<int>(n + 1));
  for (int i = 0; i <= n; i++)
    for (int j = 0; j < i; j++) dp[i][j] = -1;
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      int index = s[i - 1] - 'a';
      if (count2[index][j] < count1[index][i])
        dp[i][j] = -1;
      else if (index == t[j - 1] - 'a') {
        dp[i][j] = dp[i - 1][j - 1];
      } else {
        dp[i][j] = (dp[i - 1][j] == -1) ? dp[i - 1][j] : dp[i - 1][j] + 1;
        if (dp[i][j - 1] > -1) {
          if (dp[i][j] == -1 || dp[i][j] > dp[i][j - 1]) {
            dp[i][j] = dp[i][j - 1];
          }
        }
      }
    }
  }
  cout << dp[n][n] << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int T;
  T = 1;
  cin >> T;
  for (int tt = 1; tt <= T; tt++) {
    TestCase();
  }
}
