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")
using namespace std;
int cnts[2005][30], cntt[2005][30];
char s[2005], t[2005];
int dp[2005][2005], n;
void solve() {
  scanf("%d%s%s", &n, s + 1, t + 1);
  memset(cnts[n + 1], 0, sizeof(cnts[n + 1]));
  memset(cntt[n + 1], 0, sizeof(cntt[n + 1]));
  for (int i = n; i >= 1; i--) {
    for (int j = 0; j <= 25; 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 <= 25; i++) {
    if (cnts[1][i] != cntt[1][i]) {
      puts("-1");
      return;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
      if (s[i] == t[j]) {
        bool f = 1;
        for (int k = 0; k <= 25; k++)
          if (cnts[i][k] < cntt[j][k]) {
            f = 0;
            break;
          }
        if (f) dp[i][j] = dp[i - 1][j - 1] + 1;
      }
    }
  }
  printf("%d\n", n - dp[n][n]);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) solve();
  return 0;
}
