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::chrono;
using namespace std;
const long long mod = 1000000007;
const double pi = 3.14159265359;
template <typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
  cerr << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char *names, Arg1 &&arg1, Args &&...args) {
  const char *comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
string s, g;
long long n;
map<char, long long> mp;
long long dp[2002][2002];
long long fun(long long i, long long j) {
  long long ret = 0;
  if (i == -1) {
    for (auto it : mp) ret += it.second;
    return ret;
  } else if (dp[i][j] != -1)
    return dp[i][j];
  else if (s[i] == g[j])
    return dp[i][j] = fun(i - 1, j - 1);
  else {
    mp[s[i]]++, ret = fun(i - 1, j), mp[s[i]]--;
    if (mp[g[j]] > 0) mp[g[j]]--, ret = min(ret, fun(i, j - 1) + 1), mp[g[j]]++;
    return dp[i][j] = ret;
  }
}
int32_t main() {
  std::ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  ;
  long long t;
  cin >> t;
  while (t--) {
    cin >> n >> s >> g;
    mp.clear();
    bool possi = 1;
    vector<long long> alp(26);
    for (long long i = 0; i <= n; i++)
      for (long long j = 0; j <= n; j++) dp[i][j] = -1;
    for (long long i = 0; i < n; i++) alp[s[i] - 'a']++, alp[g[i] - 'a']--;
    for (long long i = 0; i < 26; i++)
      if (alp[i] != 0) possi = 0;
    if (possi)
      cout << fun(n - 1, n - 1) << endl;
    else
      cout << -1 << endl;
  }
}
