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 double eps = 1e-9;
const int inf = 2000000000;
const long long infLL = 9000000000000000000;
template <typename first, typename second>
ostream& operator<<(ostream& os, const pair<first, second>& p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "{";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ",";
    os << *it;
  }
  return os << "]";
}
template <typename T>
ostream& operator<<(ostream& os, const multiset<T>& v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename first, typename second>
ostream& operator<<(ostream& os, const map<first, second>& v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
void faltu() { cerr << '\n'; }
template <typename T>
void faltu(T a[], int n) {
  for (int i = 0; i < n; ++i) cerr << a[i] << ' ';
  cerr << '\n';
}
template <typename T, typename... hello>
void faltu(T arg, const hello&... rest) {
  cerr << arg << ' ';
  faltu(rest...);
}
int n;
string s;
string t;
int dp[2005][2005];
int c[30];
int d[30];
int hmm1[2005][30];
int hmm2[2005][30];
bool ul[2005][2005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int T;
  cin >> T;
  while (T--) {
    memset(c, 0, sizeof(c));
    memset(d, 0, sizeof(d));
    cin >> n;
    cin >> s;
    cin >> t;
    for (int i = 0; i < n; ++i) {
      ++c[s[i] - 'a'];
      ++d[t[i] - 'a'];
    }
    bool ok = 0;
    for (int i = 0; i < 30; ++i) {
      if (c[i] != d[i]) {
        cout << -1 << '\n';
        ok = 1;
        break;
      }
    }
    if (ok) continue;
    swap(s, t);
    for (int i = 0; i < n + 3; ++i)
      for (int j = 0; j < n + 3; ++j) dp[i][j] = 0, ul[i][j] = 0;
    for (int i = 0; i < n + 3; ++i)
      for (int j = 0; j < 30; ++j) hmm1[i][j] = hmm2[i][j] = 0;
    for (int i = n - 1; i >= 0; --i) {
      ++hmm1[i][s[i] - 'a'];
      ++hmm2[i][t[i] - 'a'];
      for (int j = 0; j < 26; ++j) {
        hmm1[i][j] += hmm1[i + 1][j];
        hmm2[i][j] += hmm2[i + 1][j];
      }
    }
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        bool ok = 0;
        for (int k = 0; k < 26; ++k) {
          if (hmm2[j][k] < hmm1[i][k]) {
            ok = 1;
            break;
          }
        }
        if (!ok) ul[i][j] = 1;
      }
    }
    for (int i = 0; i <= n; ++i) ul[n][i] = 1;
    for (int i = 0; i < n; ++i) {
      if (s[0] == t[i] && ul[1][i + 1]) dp[0][i] = 1;
    }
    int ans = 0;
    for (int i = 1; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        if (!ul[i + 1][j + 1]) continue;
        if (s[i] == t[j]) {
          dp[i][j] = 1;
          if (j) dp[i][j] += dp[i - 1][j - 1];
        }
        if (j) dp[i][j] = max(dp[i][j], dp[i][j - 1]);
        dp[i][j] = max(dp[i][j], dp[i - 1][j]);
      }
    }
    ans = dp[n - 1][n - 1];
    cout << n - ans << '\n';
  }
}
