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 s, p;
int dp[2005][2005];
bool vis[2005][2005];
int have[2005][26], need[2005][26];
int n;
int rec(int i, int j) {
  if (i == 0) return 0;
  if (vis[i][j]) return dp[i][j];
  vis[i][j] = 1;
  if (s[i - 1] == p[j - 1]) return dp[i][j] = rec(i - 1, j - 1);
  dp[i][j] = rec(i - 1, j) + 1;
  if (have[n - i][p[j - 1] - 'a'] - need[n - j][p[j - 1] - 'a'] > 0) {
    dp[i][j] = min(dp[i][j], rec(i, j - 1));
  }
  return dp[i][j];
}
int main() {
  int i, j, t;
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    s.clear();
    p.clear();
    for (i = 0; i < n + 1; i++) {
      for (j = 0; j < n + 1; j++) vis[i][j] = 0;
    }
    cin >> n;
    cin >> s >> p;
    string a = s, b = p;
    sort(a.begin(), a.end());
    sort(b.begin(), b.end());
    if (a != b) {
      cout << -1 << '\n';
      continue;
    }
    for (i = 1; i < n + 1; i++) {
      for (j = 0; j < 26; j++) {
        need[i][j] = need[i - 1][j];
        have[i][j] = have[i - 1][j];
        if (s[n - i] - 'a' == j) have[i][j]++;
        if (p[n - i] - 'a' == j) need[i][j]++;
      }
    }
    cout << rec(n, n) << '\n';
  }
  return 0;
}
