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 N = 1e6 + 10;
const int mod = 1e9 + 7;
int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + c - '0', c = getchar();
  return x * f;
}
char s[N], t[N];
int main() {
  int T = read();
  while (T--) {
    int n = read();
    scanf("%s", s + 1);
    scanf("%s", t + 1);
    vector<vector<int> > dp(n + 1, vector<int>(n + 1));
    vector<vector<int> > sc(n + 2, vector<int>(26));
    vector<vector<int> > tc(n + 2, vector<int>(26));
    for (int i = n; i >= 1; i--) {
      sc[i][s[i] - 'a']++, tc[i][t[i] - 'a']++;
      for (int j = 0; j < 26; j++)
        sc[i][j] += sc[i + 1][j], tc[i][j] += tc[i + 1][j];
    }
    int flag = 1;
    for (int j = 0; j < 26; j++)
      if (sc[1][j] != tc[1][j]) flag = 0;
    if (!flag) {
      puts("-1");
      continue;
    }
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
        dp[i][j] = max(dp[i][j - 1], dp[i - 1][j]);
        if (s[i] == t[j]) {
          int ok = 1;
          for (int t = 0; t < 26; t++)
            if (sc[i][t] < tc[j][t]) ok = 0;
          if (ok) dp[i][j] = max(dp[i - 1][j - 1] + 1, dp[i][j]);
        }
      }
    }
    cout << n - dp[n][n] << endl;
  }
}
