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;
void solve() {
  int n;
  cin >> n;
  string a, b;
  cin >> a >> b;
  vector<vector<int>> aCount(n + 1, vector<int>(26, 0)),
      bCount(n + 1, vector<int>(26, 0));
  for (int i = 0; i < n; i++) {
    a[i] -= 'a';
    b[i] -= 'a';
    aCount[i + 1] = aCount[i];
    bCount[i + 1] = bCount[i];
    aCount[i + 1][a[i]]++;
    bCount[i + 1][b[i]]++;
  }
  for (int i = 0; i < 26; i++) {
    if (aCount.back()[i] != bCount.back()[i]) {
      cout << -1 << endl;
      return;
    }
  }
  int large = 1 << 30;
  vector<vector<int>> dp(n + 1, vector<int>(n + 1, large));
  for (int i = n; i >= 0; i--) {
    for (int j = n; j >= i; j--) {
      if (j == n) {
        dp[i][n] = 0;
        continue;
      }
      char c = b[j];
      int op = dp[i + 1][j];
      if (a[i] == b[j]) op = min(dp[i + 1][j + 1], op);
      if ((aCount[n][c] - aCount[i][c]) >= (bCount[n][c] - bCount[j][c]))
        op = min(dp[i][j + 1] + 1, op);
      dp[i][j] = op;
    }
  }
  cout << dp[0][0] << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
