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 = 2005;
const long long MAX = 1e18;
string s, t;
long long tc, n, dp[N][N], suff1[26][N], suff2[26][N];
long long solve(int i, int j) {
  if (!j) return 0;
  if (~dp[i][j]) return dp[i][j];
  long long ans = MAX;
  if (i)
    ans = min(solve(i - 1, j),
              (s[i - 1] == t[j - 1] ? solve(i - 1, j - 1) : ans));
  if (suff1[t[j - 1] - 'a'][i] > suff2[t[j - 1] - 'a'][j])
    ans = min(ans, solve(i, j - 1) + 1);
  return dp[i][j] = ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> tc;
  while (tc--) {
    cin >> n >> s >> t;
    for (int i = 0; i < n + 5; ++i)
      for (int j = 0; j < n + 5; ++j) dp[i][j] = -1;
    for (int i = 0; i < 26; ++i)
      for (int j = 0; j <= n; ++j) suff1[i][j] = suff2[i][j] = 0;
    for (int i = 0; i < 26; ++i)
      for (int j = n - 1; j >= 0; --j)
        suff1[i][j] = suff1[i][j + 1] + (s[j] - 'a' == i),
        suff2[i][j] = suff2[i][j + 1] + (t[j] - 'a' == i);
    long long ans = solve(n, n);
    cout << (ans >= MAX ? -1 : ans) << '\n';
  }
  return 0;
}
