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;
vector<vector<int>> dp;
vector<int> freeChars;
string s, t;
int solve(int poss, int post) {
  if (poss < 0) return 0;
  if (dp[poss][post] != -1) return dp[poss][post];
  int ret = 1e9;
  if (s[poss] == t[post]) ret = solve(poss - 1, post - 1);
  if (freeChars[t[post] - 'a'] > 0) {
    freeChars[t[post] - 'a']--;
    ret = min(ret, solve(poss, post - 1));
    freeChars[t[post] - 'a']++;
  }
  freeChars[s[poss] - 'a']++;
  ret = min(ret, 1 + solve(poss - 1, post));
  freeChars[s[poss] - 'a']--;
  return dp[poss][post] = ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int q;
  cin >> q;
  for (int T = (0); T < (q); T++) {
    int n;
    cin >> n >> s >> t;
    vector<int> cants(26, 0), cantt(26, 0);
    for (int i = (0); i < (n); i++) {
      cants[s[i] - 'a']++;
      cantt[t[i] - 'a']++;
    }
    bool ok = true;
    for (int i = (0); i < (26); i++)
      if (cants[i] != cantt[i]) ok = false;
    if (!ok) {
      cout << "-1\n";
      continue;
    }
    dp.resize(n, vector<int>(n, -1));
    freeChars.resize(26, 0);
    cout << solve(n - 1, n - 1) << '\n';
    dp.clear();
    freeChars.clear();
  }
  return 0;
}
