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;
template <typename Tp>
inline void outarr(Tp _begin, Tp _end, const char* _delim = " ") {
  for (Tp current = _begin; current != _end; ++current) {
    std::cout << *current << _delim;
  }
  std::cout << '\n';
}
using ll = long long;
using pii = std::pair<int, int>;
constexpr int INF = 0x3f3f3f3f;
constexpr int MOD = static_cast<const int>(1e9 + 7);
vector<vector<int>> Parse(string_view s) {
  vector<vector<int>> suff(s.size() + 1, vector<int>(26, 0));
  for (int i = static_cast<int>((s).size()) - 1; i >= 0; --i) {
    suff[i] = suff[i + 1];
    ++suff[i][s[i] - 'a'];
  }
  return suff;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int test_cases;
  cin >> test_cases;
  string s, t;
  while (test_cases--) {
    int n;
    cin >> n >> t >> s;
    auto suff_s = Parse(s);
    auto suff_t = Parse(t);
    if (suff_s[0] != suff_t[0]) {
      cout << -1 << "\n";
      continue;
    }
    vector<vector<int>> dp(n + 1, vector<int>(n + 1, 0));
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= i; ++j) {
        dp[i][j] = dp[i][j - 1];
        if (s[i - 1] == t[j - 1]) {
          dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + 1);
        } else if (suff_t[j][s[i - 1] - 'a'] > suff_s[i][s[i - 1] - 'a']) {
          dp[i][j] = max(dp[i][j], dp[i - 1][j]);
        }
      }
    }
    cout << n - dp[n][n] << "\n";
  }
  return 0;
}
