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 = 2020;
int dp[MAXN][MAXN];
int allA[MAXN][30];
int allB[MAXN][30];
char A[MAXN];
char B[MAXN];
int n;
int main(void) {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int testCase = 0;
  cin >> testCase;
  while (testCase--) {
    cin >> n;
    cin >> (A + 1);
    cin >> (B + 1);
    for (int i = 1; i <= n; ++i) A[i] -= 'a', B[i] -= 'a';
    for (int i = 1; i <= n; ++i)
      for (int j = 0; j <= 25; ++j)
        allA[i][j] = allA[i - 1][j] + (int(A[i]) == j),
        allB[i][j] = allB[i - 1][j] + (int(B[i]) == j);
    bool nxt = 0;
    for (int i = 0; i <= 25; ++i)
      if (allA[n][i] != allB[n][i]) {
        nxt = 1, cout << "-1\n";
        break;
      }
    if (nxt) continue;
    for (int i = 1; i <= n; ++i)
      for (int j = i; j <= n; ++j) {
        dp[i][j] = dp[i - 1][j] + 1;
        if (A[i] == B[j]) dp[i][j] = min(dp[i][j], dp[i - 1][j - 1]);
        if (allA[i][B[j]] < allB[j][B[j]])
          dp[i][j] = min(dp[i][j], dp[i][j - 1]);
      }
    cout << dp[n][n] << '\n';
  }
  return 0;
}
