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;
using ll = long long;
const int N = 2005;
int n;
char s[N], t[N];
int dp[N][N];
int mark[N][N];
int pass;
int cnta[N][26], cntb[N][26];
int f(int i, int j) {
  if (i == n or j == n) return 0;
  int &ans = dp[i][j];
  if (mark[i][j] == pass) return ans;
  mark[i][j] = pass;
  ans = 1 + f(i, j + 1);
  if (s[i] == t[j]) {
    ans = min(ans, f(i + 1, j + 1));
  } else {
    int c = int(s[i] - 'a');
    if (cnta[i][c] > cntb[j][c]) {
      ans = min(ans, f(i + 1, j));
    }
  }
  return ans;
}
int solve() {
  memset(cnta, 0, sizeof cnta);
  memset(cntb, 0, sizeof cntb);
  for (int i = n - 1; i >= 0; i--) {
    for (int j = 0; j < 26; j++) {
      cnta[i][j] = cnta[i + 1][j];
      cntb[i][j] = cntb[i + 1][j];
    }
    cnta[i][(int(s[i] - 'a'))]++;
    cntb[i][(int(t[i] - 'a'))]++;
  }
  for (int j = 0; j < 26; j++) {
    if (cnta[0][j] != cntb[0][j]) {
      return -1;
    }
  }
  pass++;
  int tmp = f(0, 0);
  return tmp;
}
int main() {
  int tc;
  scanf("%d", &tc);
  while (tc--) {
    scanf("%d %s %s", &n, s, t);
    printf("%d\n", solve());
  }
}
