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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T, n;
  string s, t;
  cin >> T;
  while (T--) {
    cin >> n >> s >> t;
    reverse(s.begin(), s.end());
    reverse(t.begin(), t.end());
    s = '.' + s;
    t = '.' + t;
    vector<vector<int> > dp(n + 1, vector<int>(n + 1));
    vector<vector<int> > pre_s(n + 2, vector<int>(26)),
        pre_t(n + 2, vector<int>(26));
    for (int i = 1; i <= n; i++) {
      for (int k = 0; k <= 25; k++)
        pre_s[i][k] = pre_s[i - 1][k], pre_t[i][k] = pre_t[i - 1][k];
      ++pre_s[i][s[i] - 'a'];
      ++pre_t[i][t[i] - 'a'];
    }
    bool ok = 1;
    for (int i = 0; i <= 25; i++)
      if (pre_s[n][i] != pre_t[n][i]) ok = 0;
    if (!ok) {
      cout << -1 << endl;
      continue;
    }
    for (int i = 0; i <= n; i++) dp[i][0] = i;
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= i; j++) {
        bool ok = 1;
        for (int k = 0; k <= 25; k++)
          if (pre_s[i][k] < pre_t[j][k]) ok = 0;
        if (!ok) continue;
        dp[i][j] = dp[i][j - 1];
        ok = 1;
        for (int k = 0; k <= 25; k++)
          if (pre_s[i - 1][k] < pre_t[j][k]) ok = 0;
        if (ok) dp[i][j] = min(dp[i][j], dp[i - 1][j] + 1);
        if (s[i] == t[j]) dp[i][j] = min(dp[i][j], dp[i - 1][j - 1]);
      }
    cout << dp[n][n] << endl;
  }
}
