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>
#pragma GCC optimize "03"
using namespace std;
const long long int inf = 1e15;
const long long int MOD = 1e9 + 7;
const long long int N = 2005;
long long int dp[N][N];
long long int cnt1[N][27], cnt2[N][27];
string s, t;
long long int solve(long long int i, long long int j) {
  if (i == -1 || j == -1) return 0;
  if (dp[i][j] != -1) return dp[i][j];
  long long int ans = inf;
  if (s[i] == t[j]) ans = min(ans, solve(i - 1, j - 1));
  if (cnt1[i + 1][t[j] - 'a'] > cnt2[j + 1][t[j] - 'a'])
    ans = min(ans, solve(i, j - 1));
  ans = min(ans, solve(i - 1, j) + 1);
  dp[i][j] = ans;
  return ans;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long int T;
  cin >> T;
  while (T--) {
    long long int n;
    cin >> n;
    cin >> s >> t;
    for (long long int i = 0; i <= n; i++) {
      for (long long int j = 0; j <= n; j++) {
        dp[i][j] = -1;
      }
      for (long long int j = 0; j < 26; j++) cnt1[i][j] = cnt2[i][j] = 0;
    }
    cnt1[n - 1][s[n - 1] - 'a']++;
    cnt2[n - 1][t[n - 1] - 'a']++;
    for (long long int i = n - 2; i >= 0; i--) {
      for (long long int j = 0; j < 26; j++) {
        cnt1[i][j] = cnt1[i + 1][j];
        cnt2[i][j] = cnt2[i + 1][j];
      }
      cnt1[i][s[i] - 'a']++;
      cnt2[i][t[i] - 'a']++;
    }
    long long int flag = 1;
    for (long long int i = 0; i < 26; i++) {
      if (cnt1[0][i] != cnt2[0][i]) flag = 0;
    }
    if (flag)
      cout << solve(n - 1, n - 1) << '\n';
    else
      cout << -1 << '\n';
  }
  return 0;
}
