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;
template <class T>
void dbs(string str, T t) {
  cerr << str << " : " << t << "\n";
}
template <class T, class... S>
void dbs(string str, T t, S... s) {
  int idx = str.find(',');
  cerr << str.substr(0, idx) << " : " << t << ",";
  dbs(str.substr(idx + 1), s...);
}
template <class S, class T>
ostream& operator<<(ostream& os, const pair<S, T>& p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <class T>
void debug(T a, T b) {
  cerr << "[";
  for (T i = a; i != b; ++i) {
    if (i != a) cerr << ", ";
    cerr << *i;
  }
  cerr << "]\n";
}
const int N = 2020;
int dp[N][N];
bool ok[N][N];
string s, t;
int go(int i, int j) {
  if (i == s.length() || j == s.length()) {
    return 0;
  }
  int& ret = dp[i][j];
  if (~ret) {
    return ret;
  }
  ret = max(go(i + 1, j), go(i, j + 1));
  if (ok[i][j] and s[i] == t[j]) {
    ret = max(ret, 1 + go(i + 1, j + 1));
  }
  return ret;
}
bitset<N> bits[N];
int a[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int q;
  cin >> q;
  while (q--) {
    int n;
    cin >> n;
    cin >> s;
    cin >> t;
    vector<int> freqS(26, 0);
    for (int i = n - 1; i >= 0; --i) {
      ++freqS[s[i] - 'a'];
      vector<int> freqT(26, 0);
      int okCount = 26;
      for (int j = n - 1; j >= 0; --j) {
        int charIdx = t[j] - 'a';
        okCount -= freqS[charIdx] >= freqT[charIdx];
        ++freqT[charIdx];
        okCount += freqS[charIdx] >= freqT[charIdx];
        ok[i][j] = (okCount == 26);
      }
    }
    if (!ok[0][0]) {
      cout << "-1\n";
      continue;
    }
    for (int i = 0; i <= n; ++i) {
      for (int j = 0; j <= n; ++j) {
        dp[i][j] = -1;
      }
    }
    cout << n - go(0, 0) << "\n";
  }
  return 0;
}
