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 long long N = 3000;
vector<vector<long long>> mem(3000, vector<long long>(3000, -1));
string s, t;
vector<vector<long long>> sufs(26, vector<long long>(3000)), suft = sufs;
long long cnt = 0;
long long sol(long long i, long long j) {
  if (j == 0) return 0;
  long long &ans = mem[i][j];
  if (ans != -1) return ans;
  ans = 1e9;
  cnt++;
  if (i > 0) {
    ans = 1 + sol(i - 1, j);
    if (s[i - 1] == t[j - 1]) {
      ans = min(ans, sol(i - 1, j - 1));
    }
  }
  long long c = t[j - 1] - 'a';
  if (sufs[c][i + 1] > suft[c][j + 1]) {
    ans = min(ans, sol(i, j - 1));
  }
  return ans;
}
signed main() {
  long long tt;
  cin >> tt;
  while (tt--) {
    long long n;
    cnt = 0;
    cin >> n >> s >> t;
    for (long long i = 0; i <= n; i++) {
      for (long long j = 0; j <= n; ++j) {
        mem[i][j] = -1;
      }
    }
    for (long long i = 0; i < 26; ++i) {
      for (long long j = 0; j <= n + 1; ++j) {
        sufs[i][j] = 0;
        suft[i][j] = 0;
      }
    }
    for (long long i = n; i >= 1; --i) {
      for (long long j = 0; j < 26; ++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]++;
    }
    long long res = sol(n, n);
    cout << (res > 1e9 ? -1 : res) << '\n';
  }
  return 0;
}
