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>
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector")
using namespace std;
const int N = 2005, inf = 1e9 + 7;
int dp[N][N];
int cnt[2][26][N];
string second[2];
int n;
template <class T>
void umin(T &a, T b) {
  a = min(a, b);
}
int main() {
  cin.tie(0)->sync_with_stdio(0), cout.tie(0);
  cin >> n;
  while (cin >> n >> second[1] >> second[0]) {
    for (auto i = (0) - ((0) > (n)); i != n - ((0) > (n));
         i += 1 - 2 * ((0) > (n)))
      for (auto c = (0) - ((0) > (26)); c != 26 - ((0) > (26));
           c += 1 - 2 * ((0) > (26))) {
        cnt[0][c][i + 1] = cnt[0][c][i] + (second[0][i] - 'a' == c);
        cnt[1][c][i + 1] = cnt[1][c][i] + (second[1][i] - 'a' == c);
      }
    for (auto i = (0) - ((0) > (n + 1)); i != n + 1 - ((0) > (n + 1));
         i += 1 - 2 * ((0) > (n + 1)))
      for (auto j = (0) - ((0) > (n + 1)); j != n + 1 - ((0) > (n + 1));
           j += 1 - 2 * ((0) > (n + 1)))
        dp[i][j] = inf;
    dp[0][0] = 0;
    for (auto i = (0) - ((0) > (n + 1)); i != n + 1 - ((0) > (n + 1));
         i += 1 - 2 * ((0) > (n + 1)))
      for (auto j = (0) - ((0) > (i + 1)); j != i + 1 - ((0) > (i + 1));
           j += 1 - 2 * ((0) > (i + 1))) {
        if (i < n && j < n && second[0][i] == second[1][j])
          umin(dp[i + 1][j + 1], dp[i][j]);
        if (i < n) umin(dp[i + 1][j], dp[i][j] + 1);
        if (j < n) {
          int c = second[1][j] - 'a';
          if (cnt[0][c][i] >= cnt[1][c][j + 1]) umin(dp[i][j + 1], dp[i][j]);
        }
      }
    if (dp[n][n] >= inf) dp[n][n] = -1;
    cout << dp[n][n] << '\n';
  }
  return 0;
}
