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;
int main() {
  int T;
  cin >> T;
  while (T--) {
    int n;
    string s, t;
    cin >> n >> s >> t;
    string s1 = s, s2 = t;
    sort(s1.begin(), s1.end());
    sort(s2.begin(), s2.end());
    if (s1 != s2)
      printf("-1\n");
    else {
      int cnt1[n + 1][26], cnt2[n + 1][26], dp[n + 1][n + 1];
      memset(cnt1, 0, sizeof cnt1);
      memset(cnt2, 0, sizeof cnt2);
      memset(dp, 0, sizeof dp);
      int l = 0, r = n - 1;
      while (s[l] == t[l] && l < n) l++;
      while (s[r] == t[r] && r >= 0) r--;
      for (int i = r; i >= l; i--) {
        cnt1[i][s[i] - 'a']++;
        cnt2[i][t[i] - 'a']++;
        for (int j = 0; j < 26; j++) {
          cnt1[i][j] += cnt1[i + 1][j];
          cnt2[i][j] += cnt2[i + 1][j];
        }
      }
      if (l <= r) {
        for (int i = l; i <= r; i++) {
          for (int j = l; j <= r; j++) {
            if (i > 0) dp[i][j] = dp[i - 1][j] + 1;
            if (j > 0 && cnt1[i + 1][t[j] - 'a'] > cnt2[j + 1][t[j] - 'a'])
              dp[i][j] = min(dp[i][j], dp[i][j - 1]);
            if (i > 0 && j > 0 && s[i] == t[j])
              dp[i][j] = min(dp[i][j], dp[i - 1][j - 1]);
          }
        }
        printf("%d\n", dp[r][r]);
      } else
        printf("0\n");
    }
  }
}
