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 double eps = 1e-9;
const int inf = (1 << 30) - 1;
const long long inf64 = ((long long)1 << 62) - 1;
const long double pi = acos(-1);
template <class T>
T sqr(T x) {
  return x * x;
}
template <class T>
T abs(T x) {
  return x < 0 ? -x : x;
}
int main() {
  int tt;
  scanf("%d", &tt);
  while (tt--) {
    int m;
    scanf("%d", &m);
    string s1, s2;
    cin >> s1 >> s2;
    int n = ((int)(s1).size());
    vector<int> a1(n);
    vector<int> a2(n);
    for (int i = 0; i < n; ++i) {
      a1[i] = s1[i] - 'a';
      a2[i] = s2[i] - 'a';
    }
    vector<vector<int> > num1(26, vector<int>(n + 1, 0));
    vector<vector<int> > num2(26, vector<int>(n + 1, 0));
    for (int i = 0; i < 26; ++i) {
      num1[i][n] = 0;
      num2[i][n] = 0;
    }
    for (int i = n - 1; i >= 0; --i) {
      for (int j = 0; j < 26; ++j) {
        num1[j][i] = num1[j][i + 1];
        num2[j][i] = num2[j][i + 1];
      }
      ++num1[a1[i]][i];
      ++num2[a2[i]][i];
    }
    bool key = true;
    for (int i = 0; i < 26; ++i) {
      if (num1[i][0] != num2[i][0]) {
        key = false;
        break;
      }
    }
    if (!key) {
      cout << -1 << endl;
      continue;
    }
    vector<vector<int> > res(n + 1, vector<int>(n + 1));
    res[n][n] = 0;
    for (int i = 0; i < n; ++i) {
      res[n][i] = inf;
    }
    for (int i = n - 1; i >= 0; --i) {
      res[i][n] = 0;
      for (int j = n - 1; j >= 0; --j) {
        res[i][j] = inf;
        bool k1 = true;
        for (int k = 0; k < 26; ++k) {
          if (num1[k][i] < num2[k][j]) {
            k1 = false;
            break;
          }
        }
        if (!k1) {
          continue;
        }
        if (a1[i] == a2[j]) {
          res[i][j] = min(res[i][j], res[i + 1][j + 1]);
        }
        res[i][j] = min(res[i][j], res[i + 1][j]);
        if (num1[a2[j]][i] >= num2[a2[j]][j])
          res[i][j] = min(res[i][j], res[i][j + 1] + 1);
      }
    }
    cout << res[0][0] << endl;
  }
  return 0;
}
