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 T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  for (typename vector<T>::const_iterator vi = v.begin(); vi != v.end(); ++vi) {
    if (vi != v.begin()) os << ", ";
    os << *vi;
  }
  os << "}";
  return os;
}
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  os << '(' << p.first << ", " << p.second << ')';
  return os;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long T, N, M, K;
string s, t;
int dp[2010][2010], suf[2010][26], suf2[2010][26];
int solve(int i, int j) {
  if (j == 0) return 0;
  if (dp[i][j] != -1) return dp[i][j];
  int ans = 0x3f3f3f3f;
  if (i > 0) {
    ans = min(ans, 1 + solve(i - 1, j));
    if (s[i - 1] == t[j - 1]) ans = min(ans, solve(i - 1, j - 1));
  }
  int c = t[j - 1] - 'a';
  if (suf[i + 1][c] > suf2[j + 1][c]) ans = min(ans, solve(i, j - 1));
  dp[i][j] = ans;
  return dp[i][j];
}
void solve() {
  int ans = solve(N, N);
  if (ans == 0x3f3f3f3f)
    cout << -1 << "\n";
  else
    cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> T;
  while (T--) {
    cin >> N;
    cin >> s >> t;
    for (int i = 0; i <= N; i++) {
      for (int j = 0; j <= N; j++) {
        dp[i][j] = -1;
      }
      for (int j = 0; j < 26; j++) {
        suf[i][j] = -1;
        suf2[i][j] = -1;
      }
    }
    for (int i = N; i >= 1; i--) {
      for (int j = 0; j < 26; j++) {
        suf[i][j] = suf[i + 1][j];
        suf2[i][j] = suf2[i + 1][j];
      }
      suf[i][(int)s[i - 1] - 'a']++;
      suf2[i][(int)t[i - 1] - 'a']++;
    }
    solve();
  }
  return 0;
}
