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;
string a, b;
const int INF = 1e9;
int dp[2005][2005], chk[2005][2005];
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  int t, cnt[26] = {};
  cin >> t;
  while (t--) {
    int n;
    cin >> n >> a >> b;
    for (int i = 0; i <= ((int)b.size()); ++i)
      for (int j = 0; j <= ((int)a.size()); ++j) dp[i][j] = INF, chk[i][j] = 0;
    fill(dp[((int)b.size())], dp[((int)b.size())] + ((int)a.size()) + 1, 0);
    fill(chk[((int)b.size())], chk[((int)b.size())] + ((int)a.size()) + 1, 1);
    fill(cnt, cnt + 26, 0);
    for (int i = ((int)b.size()) - 1; i >= 0; --i) {
      ++cnt[b[i] - 'a'];
      int fin = 0, cnt2[26] = {};
      for (int j = 0; j < 26; ++j)
        if (cnt[j] == 0) ++fin;
      for (int j = ((int)a.size()) - 1; j >= 0; --j) {
        ++cnt2[a[j] - 'a'];
        if (cnt2[a[j] - 'a'] == cnt[a[j] - 'a']) ++fin;
        if (fin == 26) chk[i][j] = 1;
      }
    }
    if (chk[0][0] != 1) {
      cout << "-1\n";
      continue;
    }
    for (int i = ((int)b.size()) - 1; i >= 0; --i)
      for (int j = ((int)a.size()) - 1; j >= 0; --j) {
        dp[i][j] = dp[i][j + 1];
        if (b[i] == a[j]) dp[i][j] = min(dp[i][j], dp[i + 1][j + 1]);
        if (chk[i][j]) dp[i][j] = min(dp[i][j], dp[i + 1][j] + 1);
      }
    cout << dp[0][0] << "\n";
  }
}
