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 maxn = 2100;
int cnt[2][maxn][26];
int dp[maxn][maxn];
string s[2];
int solve() {
  int n = s[0].size();
  for (int i = 0; i < (int)(2); ++i)
    for (int j = 0; j < (int)(n + 1); ++j)
      for (int k = 0; k < (int)(26); ++k) cnt[i][j][k] = 0;
  for (int i = 0; i < (int)(2); ++i)
    for (int j = 0; j < (int)(n); ++j)
      for (int k = 0; k < (int)(26); ++k)
        cnt[i][j + 1][k] = cnt[i][j][k] + (s[i][j] == 'a' + k);
  bool ok = true;
  for (int k = 0; k < (int)(26); ++k) ok &= cnt[0][n][k] == cnt[1][n][k];
  if (!ok) return -1;
  for (int i = 0; i < (int)(n + 1); ++i)
    for (int j = 0; j < (int)(n + 1); ++j) dp[i][j] = 0;
  for (int i = 0; i < (int)(n + 1); ++i)
    for (int j = 0; j < (int)(n + 1); ++j) {
      dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
      dp[i][j + 1] = max(dp[i][j + 1], dp[i][j]);
      if (s[0][i] == s[1][j]) {
        bool ok = true;
        for (int k = 0; k < (int)(26); ++k) ok &= cnt[0][i][k] <= cnt[1][j][k];
        if (ok) dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j] + 1);
      }
    }
  return n - dp[n][n];
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.precision(10);
  cout << fixed;
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n >> s[0] >> s[1];
    cout << solve() << '\n';
  }
  return 0;
}
