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;
long long te, n, dp[2005][2005], sumt[2005][30], sums[2005][30];
string s, t;
long long calc(int p1, int p2) {
  if (p1 == t.size() && p2 == s.size()) return 0;
  if (p1 == t.size()) {
    if ((sumt[p1][s[p2] - 'a'] - sums[p2][s[p2] - 'a']) > 0)
      return 1 + dp[p1][p2 + 1];
    else
      return 3000;
  }
  long long res = dp[p1 + 1][p2];
  if ((sumt[p1][s[p2] - 'a'] - sums[p2][s[p2] - 'a']) > 0)
    res = min(res, 1 + dp[p1][p2 + 1]);
  if (t[p1] == s[p2]) res = min(res, dp[p1 + 1][p2 + 1]);
  return min(res, 3000ll);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> te;
  while (te--) {
    cin >> n >> s >> t;
    memset(sums, 0, sizeof(sums));
    memset(sumt, 0, sizeof(sumt));
    for (int i = 0; i <= t.size(); i++) {
      if (i == 0) continue;
      for (int j = 0; j < 30; j++) {
        sumt[i][j] = sumt[i - 1][j];
      }
      sumt[i][t[i - 1] - 'a']++;
    }
    for (int i = 0; i <= s.size(); i++) {
      if (i == 0) continue;
      for (int j = 0; j < 30; j++) {
        sums[i][j] = sums[i - 1][j];
      }
      sums[i][s[i - 1] - 'a']++;
    }
    for (int i = t.size(); i >= 0; i--) {
      for (int j = s.size(); j >= 0; j--) {
        dp[i][j] = calc(i, j);
      }
    }
    if (dp[0][0] == 3000)
      cout << "-1\n";
    else
      cout << dp[0][0] << "\n";
  }
  return 0;
}
