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 = 2e3 + 7;
const int inf = 1e9 + 7;
int T, n, dp[N][N], cnt[N], qzs[N][26], qzt[N][26];
char s[N], t[N];
void Min(int &x, int y) {
  if (x > y) x = y;
}
void Main() {
  fill(cnt, cnt + 26, 0);
  scanf("%d", &n);
  for (int i = 0, iE = n; i <= iE; i++)
    for (int j = 0, jE = n; j <= jE; j++) dp[i][j] = inf;
  scanf("%s%s", s + 1, t + 1);
  for (int i = 1, iE = n; i <= iE; i++) cnt[s[i] - 'a']++, cnt[t[i] - 'a']--;
  for (int i = 0, iE = 25; i <= iE; i++)
    if (cnt[i]) return puts("-1"), void();
  for (int i = 1, iE = n; i <= iE; i++)
    for (int j = 0, jE = 25; j <= jE; j++)
      qzs[i][j] = qzs[i - 1][j] + (s[i] == 'a' + j),
      qzt[i][j] = qzt[i - 1][j] + (t[i] == 'a' + j);
  for (int i = 0, iE = n; i <= iE; i++) dp[0][i] = 0;
  for (int i = 1, iE = n; i <= iE; i++)
    for (int j = i, jE = n; j <= jE; j++) {
      Min(dp[i][j], dp[i - 1][j] + 1);
      if (s[i] == t[j]) Min(dp[i][j], dp[i - 1][j - 1]);
      if (qzs[i][t[j] - 'a'] < qzt[j][t[j] - 'a']) Min(dp[i][j], dp[i][j - 1]);
    }
  printf("%d\n", dp[n][n]);
}
int main() {
  scanf("%d", &T);
  while (T--) Main();
  return 0;
}
