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 MX = 1000000;
int dp[2001][2001] = {}, cnt[26] = {};
bool pos[2001][2001] = {};
int main() {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    string s, t, a, b;
    cin >> n;
    cin >> s >> t;
    a = s;
    b = t;
    sort(a.begin(), a.end());
    sort(b.begin(), b.end());
    if (a == b) {
      for (int i = 0; i < 26; i++) {
        cnt[i] = 0;
      }
      for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
          if (s[i - 1] == t[j - 1]) {
            pos[i][j] = 1;
          } else {
            pos[i][j] = 0;
          }
        }
      }
      for (int i = 1; i <= n; i++) {
        dp[0][i] = MX;
      }
      dp[0][0] = 0;
      for (int i = 1; i <= n; i++) {
        int ar[26] = {}, mn = MX;
        cnt[s[i - 1] - 'a']++;
        for (int j = 0; j <= n; j++) {
          dp[i][j] = MX;
          if (j) {
            ar[t[j - 1] - 'a']++;
          }
          if (pos[i][j]) dp[i][j] = mn;
          if (ar[s[i - 1] - 'a'] >= cnt[s[i - 1] - 'a'])
            dp[i][j] = min(dp[i][j], dp[i - 1][j]);
          mn = min(mn + 1, dp[i - 1][j]);
        }
      }
      cout << dp[n][n] << "\n";
    } else {
      cout << "-1\n";
    }
  }
  return 0;
}
