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;
const int N = 2e3 + 5, C = 26 + 4;
int tst, n, dp[N], cnt[C];
vector<int> pos[C];
string s, t;
void read_input() {
  cin >> n >> s >> t;
  for (int i = 0; i < n; i++) pos[t[i] - 'a'].push_back(i);
}
void solve() {
  int mx = -1;
  for (auto c : s) {
    int x = c - 'a';
    if (pos[x].size() == cnt[x]) {
      cout << -1;
      return;
    }
    mx = max(mx, pos[x][cnt[x]++]);
    for (int i = n - 1, j = pos[x].size() - 1; i >= 0; i--) {
      while (~j && pos[x][j] > i) j--;
      if (~j && pos[x][j] >= mx)
        dp[i] = max(dp[i], dp[max(pos[x][j] - 1, 0)] + 1);
      else if (!~j)
        dp[i] = -n;
    }
  }
  cout << n - dp[n - 1];
}
void reset() {
  memset(dp, 0, sizeof dp);
  memset(cnt, 0, sizeof cnt);
  for (int i = 0; i < C; i++) pos[i].clear();
  cout << endl;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  for (cin >> tst; tst--; reset()) read_input(), solve();
  return 0;
}
