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 = 2020;
int n;
string s, t;
int cnt[26][2][N];
int dp[N][N];
void solve() {
  cin >> n >> s >> t;
  string ta = s, tb = t;
  sort((ta).begin(), (ta).end());
  sort((tb).begin(), (tb).end());
  if (ta != tb) return void(cout << -1 << '\n');
  for (int c = 0; c < 26; c++)
    for (int i = 0; i < 2; i++) cnt[c][i][n] = 0;
  for (int i = n - 1; i >= 0; i--) {
    for (int c = 0; c < 26; c++) {
      cnt[c][0][i] = cnt[c][0][i + 1];
      cnt[c][1][i] = cnt[c][1][i + 1];
    }
    cnt[s[i] - 'a'][0][i]++;
    cnt[t[i] - 'a'][1][i]++;
  }
  for (int i = 0; i <= n; i++) dp[0][i] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      int c1 = s[i - 1] - 'a';
      int c2 = t[j - 1] - 'a';
      dp[i][j] = 1 + dp[i - 1][j];
      if (c1 == c2) dp[i][j] = min(dp[i][j], dp[i - 1][j - 1]);
      if (cnt[c2][0][i] > cnt[c2][1][j]) dp[i][j] = min(dp[i][j], dp[i][j - 1]);
    }
  }
  cout << dp[n][n] << '\n';
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}
