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;
string a, b;
int n, q, i, j, f[2001][2001], ts[26], cnta[2001][26], cntb[2001][26];
const int oo = round(1e9);
int main() {
  cin >> q;
  while (q--) {
    cin >> n >> a >> b;
    memset(ts, 0, sizeof(ts));
    for (i = 0; i < n; i++) {
      ts[a[i] - 'a']++;
      ts[b[i] - 'a']--;
    }
    bool kt = 1;
    for (i = 0; i < 26; i++)
      if (ts[i] != 0) kt = 0;
    if (kt == 0) {
      cout << -1 << '\n';
      continue;
    }
    a = " " + a;
    b = " " + b;
    memset(cnta, 0, sizeof(cnta));
    memset(cntb, 0, sizeof(cntb));
    for (i = n; i >= 1; i--) {
      for (j = 0; j < 26; j++) {
        cnta[i][j] = cnta[i + 1][j];
        cntb[i][j] = cntb[i + 1][j];
      }
      cnta[i][a[i] - 'a']++;
      cntb[i][b[i] - 'a']++;
    }
    for (i = 0; i <= n; i++)
      for (j = 0; j <= n; j++) f[i][j] = oo;
    for (i = 0; i <= n; i++)
      for (j = 0; j <= n; j++)
        if (i == 0 && j == 0)
          f[i][j] = 0;
        else {
          if (j == 0) continue;
          int x = b[j] - 'a';
          if (cnta[i + 1][x] < cntb[j + 1][x]) continue;
          if (cnta[i + 1][x] > cntb[j + 1][x]) {
            f[i][j] = min(f[i][j], f[i][j - 1] + 1);
          }
          if (i > 0) {
            if (a[i] == b[j]) f[i][j] = min(f[i][j], f[i - 1][j - 1]);
            f[i][j] = min(f[i][j], f[i - 1][j]);
          }
        }
    cout << f[n][n] << '\n';
  }
}
