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 = 2011;
const long long oo = 1e9 + 123;
int n;
int f[N][N], cntS[N][26], cntT[N][26];
string s, t;
void testcase() {
  cin >> n >> s >> t;
  s = " " + s;
  t = " " + t;
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j < 26; ++j) cntS[i][j] = cntT[i][j] = 0;
  cntS[n][s[n] - 'a'] = cntT[n][t[n] - 'a'] = 1;
  for (int i = n - 1; i; --i) {
    for (int j = 0; j < 26; ++j) {
      cntS[i][j] = cntS[i + 1][j];
      cntT[i][j] = cntT[i + 1][j];
    }
    ++cntS[i][s[i] - 'a'];
    ++cntT[i][t[i] - 'a'];
  }
  for (int i = 0; i < 26; ++i)
    if (cntS[1][i] != cntT[1][i]) {
      cout << -1 << "\n";
      return;
    }
  for (int i = 0; i <= n + 1; ++i) {
    for (int j = 0; j <= n + 1; ++j) f[i][j] = oo;
  }
  f[n + 1][n + 1] = 0;
  for (int i = n + 1; i > 1; --i) {
    for (int j = n + 1; j >= i; --j)
      if (f[i][j] < oo) {
        if (s[i - 1] == t[j - 1]) {
          f[i - 1][j - 1] = min(f[i - 1][j - 1], f[i][j]);
        }
        f[i - 1][j] = min(f[i - 1][j], f[i][j] + 1);
        int x = t[j - 1] - 'a';
        if (cntS[i][x] - cntT[j][x] > 0) {
          f[i][j - 1] = min(f[i][j - 1], f[i][j]);
        }
      }
  }
  int res = n;
  for (int i = 1; i <= n + 1; ++i) res = min(res, f[1][i]);
  cout << res << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  int tcase = 1;
  cin >> tcase;
  for (int i = 1; i <= tcase; ++i) testcase();
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}
