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>
int main(void) {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cout.tie(nullptr);
  int testCase = 1;
  std::cin >> testCase;
  while (testCase--) {
    int n = 0;
    std::cin >> n;
    std::string S, T;
    std::cin >> S >> T;
    std::vector<int> Buk1(26, 0);
    auto Buk2 = Buk1;
    for (auto x : S) ++Buk1[x - 'a'];
    for (auto x : T) ++Buk2[x - 'a'];
    if (Buk1 != Buk2) {
      std::cout << "-1\n";
      continue;
    }
    std::vector<std::vector<int>> dp(n + 1, std::vector<int>(n + 1, -1)),
        all(n + 1, std::vector<int>(26, 0));
    auto all2 = all;
    dp[0][0] = 0;
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < 26; ++j)
        all[i + 1][j] = all[i][j] + (S[i] - 'a' == j),
                all2[i + 1][j] = all2[i][j] + (T[i] - 'a' == j);
    }
    for (int i = 0; i <= n; ++i)
      for (int j = i ? i : 1; j <= n; ++j) {
        int I = i - 1;
        int J = j - 1;
        int tmp = 1E9;
        if (i && S[I] == T[J] && dp[i - 1][j - 1] != -1) tmp = dp[i - 1][j - 1];
        if (all[i][T[J] - 'a'] < all2[j][T[J] - 'a'] && dp[i][j - 1] != -1)
          tmp = std::min(tmp, dp[i][j - 1] + 1);
        if (i && dp[i - 1][j] != -1) tmp = std::min(tmp, dp[i - 1][j] + 1);
        if (tmp != 1E9) dp[i][j] = tmp;
      }
    std::cout << (dp[n][n] / 2) << '\n';
  }
  return 0;
}
