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 target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
template <class T1, class T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
  os << "{" << p.first << "," << p.second << "}";
  return os;
}
const int N = 2e3 + 5;
const int oo = 1e9 + 7;
int n;
string s, t;
int dp[N][N];
int cnts[N][26];
int cntt[N][26];
int solve() {
  cin >> n >> s >> t;
  for (int i = 0; i <= n; i++) {
    dp[0][i] = 0;
  }
  s = "$" + s;
  t = "$" + t;
  for (int i = 0; i < 26; i++) cnts[n + 1][i] = cntt[n + 1][i] = 0;
  for (int i = n; i >= 1; i--) {
    for (int j = 0; j < 26; j++)
      cnts[i][j] = cnts[i + 1][j], cntt[i][j] = cntt[i + 1][j];
    cnts[i][s[i] - 'a']++;
    cntt[i][t[i] - 'a']++;
  }
  for (int i = 0; i < 26; i++)
    if (cnts[1][i] != cntt[1][i]) return -1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      dp[i][j] = oo;
      if (s[i] == t[j]) dp[i][j] = min(dp[i][j], dp[i - 1][j - 1]);
      dp[i][j] = min(dp[i][j], dp[i - 1][j] + 1);
      if (cnts[i + 1][t[j] - 'a'] > cntt[j + 1][t[j] - 'a'])
        dp[i][j] = min(dp[i][j], dp[i][j - 1]);
    }
  }
  return dp[n][n];
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int T, cs = 0;
  cin >> T;
  while (T--) {
    cout << solve() << "\n";
  }
}
