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 long double PI = acos(-1.0);
int dp[2005][2005];
int f[2005][26][2];
void solve() {
  int n;
  cin >> n;
  string s, t;
  cin >> s >> t;
  string st, tt;
  st = s;
  tt = t;
  sort(st.begin(), st.end());
  sort(tt.begin(), tt.end());
  if (st != tt) {
    cout << -1 << '\n';
    return;
  }
  for (int i = 0; i < n + 5; ++i) {
    for (int j = 0; j < n + 5; ++j) {
      dp[i][j] = 0;
    }
  }
  for (int i = 0; i < n + 5; ++i) {
    for (int j = 0; j < 26; ++j) {
      for (int k = 0; k < 2; ++k) {
        f[i][j][k] = 0;
      }
    }
  }
  f[0][s[0] - 'a'][0]++;
  f[0][t[0] - 'a'][1]++;
  for (int i = 1; i < n; ++i) {
    for (int j = 0; j < 26; ++j) {
      for (int k = 0; k < 2; ++k) {
        f[i][j][k] = f[i - 1][j][k];
      }
    }
    f[i][s[i] - 'a'][0]++;
    f[i][t[i] - 'a'][1]++;
  }
  for (int i = 0; i < n + 1; ++i) {
    for (int j = 0; j < n + 1; ++j) {
      dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
      dp[i][j + 1] = max(dp[i][j + 1], dp[i][j]);
      if (i < n and j < n and s[i] == t[j]) {
        int flag = 1;
        for (int k = 0; k < 26; ++k) {
          if (f[i][k][0] > f[j][k][1]) {
            flag = 0;
            break;
          }
        }
        if (flag) {
          dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j] + 1);
        }
      }
    }
  }
  cout << n - dp[n][n] << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
