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 <typename T>
using v2 = vector<vector<T>>;
template <typename T>
using v3 = vector<vector<vector<T>>>;
template <class T>
istream &operator>>(istream &is, vector<T> &v) {
  for (auto &elemnt : v) is >> elemnt;
  return is;
}
template <class T, class U>
istream &operator>>(istream &is, pair<T, U> &p) {
  is >> p.first >> p.second;
  return is;
}
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
const int MOD = 998244353;
const int INF = 1e9 + 200;
const long long INFLL = 2e18 + 200;
void solve() {
  int N;
  cin >> N;
  string S, T;
  cin >> S >> T;
  {
    string temp1 = S, temp2 = T;
    sort(temp1.begin(), temp1.end());
    sort(temp2.begin(), temp2.end());
    if (temp1 != temp2) {
      cout << -1 << endl;
      return;
    }
  }
  vector<vector<int>> cntS(N + 5, vector<int>(30, 0));
  vector<vector<int>> cntT(N + 5, vector<int>(30, 0));
  for (int i = N; i >= 1; i--) {
    for (int j = 0; j < 30; j++) {
      cntS[i][j] += cntS[i + 1][j];
      cntT[i][j] += cntT[i + 1][j];
    }
    cntS[i][S[i - 1] - 'a']++;
    cntT[i][T[i - 1] - 'a']++;
  }
  vector<vector<int>> dp(N + 1, vector<int>(N + 1, -1));
  function<int(int, int)> dfs = [&](int i, int j) {
    int &ret = dp[i][j];
    if (j == 0) return 0;
    if (ret != -1) return ret;
    ret = INF;
    if (i > 0) {
      ret = 1 + dfs(i - 1, j);
      if (S[i - 1] == T[j - 1]) {
        chmin(ret, dfs(i - 1, j - 1));
      }
    }
    int check = T[j - 1] - 'a';
    if (cntS[i + 1][check] > cntT[j + 1][check]) chmin(ret, dfs(i, j - 1));
    return ret;
  };
  int ans = dfs(N, N);
  cout << (ans >= INF ? -1 : ans) << endl;
}
int main() {
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) solve();
}
