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 + 10;
int cnts[256][N];
int cntt[256][N];
int n;
string s, t;
int dp[N][N];
int rec(int i, int j) {
  if (j > i) return 1e6;
  if (i == n) return 0;
  int &ans = dp[i][j];
  if (~ans) return ans;
  ans = 1e6;
  if (s[i] == t[j]) ans = min(ans, rec(i + 1, j + 1));
  ans = min(ans, 1 + rec(i + 1, j));
  if (s[i] != t[j] and cnts[t[j]][i] >= cntt[t[j]][j])
    ans = min(ans, rec(i, j + 1));
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  for (int tt = 0; tt < T; tt++) {
    cin >> n;
    cin >> s >> t;
    vector<int> f1(256), f2(256);
    for (int i = 0; i < (int)(s.size()); i++) f1[s[i]]++;
    for (int i = 0; i < (int)(t.size()); i++) f2[t[i]]++;
    if (f1 != f2) {
      cout << -1 << "\n";
      continue;
    }
    reverse((s).begin(), (s).end());
    reverse((t).begin(), (t).end());
    for (int i = 0; i < n; i++)
      for (int j = 0; j < 256; j++) cnts[j][i] = 0;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < 256; j++) cntt[j][i] = 0;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++) dp[i][j] = -1;
    for (int i = 0; i < n; i++) {
      cnts[s[i]][i] = 1;
      cntt[t[i]][i] = 1;
      if (i)
        for (int j = 0; j < 256; j++) {
          cnts[j][i] += cnts[j][i - 1];
          cntt[j][i] += cntt[j][i - 1];
        }
    }
    cout << rec(0, 0) << endl;
  }
}
