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 s, t;
int n;
int suf[26][2002], suf2[26][2002];
int dp[2002][2002];
int solvedp(int i, int j) {
  if (dp[i][j] != int(2e9)) return dp[i][j];
  if (!j) return dp[i][j] = 0;
  int &res = dp[i][j];
  if (i) {
    if (s[i - 1] == t[j - 1]) res = min(res, solvedp(i - 1, j - 1));
    res = min(res, 1 + solvedp(i - 1, j));
  }
  int c = t[j - 1] - 'a';
  if (suf[c][i] > suf2[c][j]) res = min(res, solvedp(i, j - 1));
  return res;
}
int main() {
  int tt;
  cin >> tt;
  for (int tc = 0; tc < tt; ++tc) {
    cin >> n >> s >> t;
    memset(suf, 0, sizeof(suf));
    memset(suf2, 0, sizeof(suf2));
    for (int i = 0; i < n + 2; ++i)
      for (int j = 0; j < n + 2; ++j) dp[i][j] = int(2e9);
    for (int i = n - 1; i >= 0; --i) {
      for (int j = 0; j < 26; ++j) {
        suf[j][i] = suf[j][i + 1];
        suf2[j][i] = suf2[j][i + 1];
      }
      ++suf[s[i] - 'a'][i];
      ++suf2[t[i] - 'a'][i];
    }
    int res = solvedp(n, n);
    if (res >= int(2e9))
      cout << -1 << endl;
    else
      cout << res << endl;
  }
}
