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 = 2005;
int dp[N][N], n;
string s, t;
int suf[N][26][2];
int solve(int p1, int p2) {
  if (p1 < 0) {
    return 0;
  }
  if (p1 > p2) {
    return 1e5;
  }
  if (dp[p1][p2] != -1) return dp[p1][p2];
  int x = t[p2] - 'a';
  if (s[p1] == t[p2]) {
    return dp[p1][p2] = solve(p1 - 1, p2 - 1);
  }
  int ans = solve(p1 - 1, p2) + 1;
  if (suf[p2 + 1][x][1] < suf[p1 + 1][x][0]) {
    ans = min(ans, solve(p1, p2 - 1));
  }
  return dp[p1][p2] = ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    cin >> n;
    cin >> 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 < 26; j++) {
        suf[i][j][0] = suf[i][j][1] = 0;
      }
    }
    for (int i = n - 1; i >= 0; i--) {
      for (int j = 0; j < 26; j++) {
        suf[i][j][0] = suf[i + 1][j][0];
        suf[i][j][1] = suf[i + 1][j][1];
      }
      suf[i][s[i] - 'a'][0]++;
      suf[i][t[i] - 'a'][1]++;
    }
    int ok = 1;
    for (int i = 0; i < 26; i++) {
      if (suf[0][i][0] != suf[0][i][1]) {
        ok = 0;
      }
    }
    if (!ok) {
      cout << "-1\n";
    } else {
      cout << solve(n - 1, n - 1) << "\n";
    }
  }
}
