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>
using vc = vector<T>;
template <class T>
using vvc = vc<vc<T>>;
template <class T>
void mkuni(vector<T>& v) {
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
}
long long rand_int(long long l, long long r) {
  static mt19937_64 gen(chrono::steady_clock::now().time_since_epoch().count());
  return uniform_int_distribution<long long>(l, r)(gen);
}
template <class T>
void print(T x, int suc = 1) {
  cout << x;
  if (suc == 1)
    cout << '\n';
  else
    cout << ' ';
}
template <class T>
void print(const vector<T>& v, int suc = 1) {
  for (int i = 0; i < v.size(); i++)
    print(v[i], i == (int)(v.size()) - 1 ? suc : 2);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    string s, t;
    cin >> s >> t;
    vc<vector<int>> ps(n + 1, vector<int>(26)), pt(n + 1, vector<int>(26));
    vc<vector<int>> dp(n + 1, vector<int>(n + 1, 1e9));
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j < 26; j++) {
        ps[i][j] = ps[i - 1][j];
        pt[i][j] = pt[i - 1][j];
      }
      ps[i][s[i - 1] - 'a']++;
      pt[i][t[i - 1] - 'a']++;
    }
    bool suc = 1;
    for (int j = 0; j < 26; j++) {
      if (ps[n][j] != pt[n][j]) suc = 0;
    }
    if (!suc)
      print(-1);
    else {
      dp[0][0] = 0;
      for (int i = 0; i <= n; i++) {
        for (int j = i; j <= n; j++) {
          if (j) dp[i][j] = min(dp[i][j], dp[i][j - 1] + 1);
          if (i && s[i - 1] == t[j - 1])
            dp[i][j] = min(dp[i][j], dp[i - 1][j - 1]);
          if (i && ps[i][s[i - 1] - 'a'] <= pt[j][s[i - 1] - 'a'])
            dp[i][j] = min(dp[i][j], dp[i - 1][j]);
        }
      }
      print(dp[n][n]);
    }
  }
}
