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;
void setIO() {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
const int N = 2e3 + 5;
const int M = 1e6 + 5;
const long long inf = 2e9;
const long long mod = 1e9 + 7;
int xyz = 1;
int n, p;
int dp[N][N];
int sufS[26][N];
int sufT[26][N];
string s;
string t;
int slv(int i, int j) {
  if (!j) return 0;
  int &ans = dp[i][j];
  if (dp[i][j] != -1) return ans;
  ans = inf;
  if (i) ans = min(ans, slv(i - 1, j) + 1);
  if (i && s[i - 1] == t[j - 1]) ans = min(ans, slv(i - 1, j - 1));
  int c = t[j - 1] - 'a';
  if (sufS[c][i + 1] > sufT[c][j + 1]) ans = min(ans, slv(i, j - 1));
  return ans;
}
void run() {
  cin >> n >> s >> t;
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= n; j++) dp[i][j] = -1;
  for (int j = 0; j <= 25; j++) sufS[j][n + 1] = sufT[j][n + 1] = 0;
  for (int i = n; i >= 1; i--) {
    for (int j = 0; j <= 25; j++) {
      sufS[j][i] = sufS[j][i + 1];
      sufT[j][i] = sufT[j][i + 1];
    }
    sufS[s[i - 1] - 'a'][i]++;
    sufT[t[i - 1] - 'a'][i]++;
  }
  int ans = slv(n, n);
  if (ans >= inf) {
    cout << "-1\n";
    return;
  }
  cout << ans << "\n";
}
int main() {
  setIO();
  cin >> xyz;
  while (xyz--) run();
  return 0;
}
