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 <class T>
bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(cin.failbit);
  int z;
  cin >> z;
  while (z--) {
    int n;
    cin >> n;
    string s, t;
    cin >> s >> t;
    string ss = s, st = t;
    sort(begin(ss), end(ss));
    sort(begin(st), end(st));
    if (ss != st) {
      cout << -1 << '\n';
      continue;
    }
    const int inf = 1e9;
    vector<vector<int> > dp(n + 1, vector<int>(n + 1, inf));
    dp[0][0] = 0;
    vector<int> cnts(26);
    for (int i = (0); i < (n + 1); ++i) {
      vector<int> cntt(26);
      int mn = inf;
      for (int j = (0); j < (n + 1); ++j) {
        if (i < n and j < n and s[i] == t[j]) ckmin(mn, dp[i][j]);
        if (i < n and j < n) ckmin(dp[i + 1][j + 1], mn);
        if (j < n) ckmin(dp[i][j + 1], dp[i][j]);
        if (i < n and cntt[s[i] - 'a'] > cnts[s[i] - 'a'])
          ckmin(dp[i + 1][j], dp[i][j] + 1);
        if (j < n) ++cntt[t[j] - 'a'];
      }
      if (i < n) ++cnts[s[i] - 'a'];
    }
    cout << dp[n][n] << '\n';
  }
}
