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 mxn = 2001, k = 26;
int n;
int a[2][mxn];
int second[2][mxn][k];
int dp[mxn][mxn];
void answer() {
  cin >> n;
  for (int i = 0; i < 2; i++)
    for (int j = 1; j <= n; j++) {
      char c;
      cin >> c;
      a[i][j] = c - 'a';
      memcpy(second[i][j], second[i][j - 1], sizeof(second[i][j]));
      second[i][j][a[i][j]]++;
    }
  if (memcmp(second[0][n], second[1][n], sizeof(second[0][n]))) {
    cout << -1 << '\n';
    return;
  }
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++) {
      dp[i][j] = dp[i - 1][j] + 1;
      if (a[0][i] == a[1][j]) dp[i][j] = min(dp[i][j], dp[i - 1][j - 1]);
      if (second[0][i][a[1][j]] < second[1][j][a[1][j]])
        dp[i][j] = min(dp[i][j], dp[i][j - 1]);
    }
  cout << dp[n][n] << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) answer();
  return 0;
}
