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>
T sqr(T x) {
  return x * x;
}
template <typename T>
T abs(T x) {
  return x < 0 ? -x : x;
}
template <typename T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <typename T>
bool chmin(T& x, const T& y) {
  if (x > y) {
    x = y;
    return true;
  }
  return false;
}
template <typename T>
bool chmax(T& x, const T& y) {
  if (x < y) {
    x = y;
    return true;
  }
  return false;
}
int main(int, char**) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    string s, t;
    cin >> s >> t;
    auto get_prefs = [&](string s) {
      auto ret = vector<vector<int>>(n + 1, vector<int>(26, 0));
      for (int i = 0; i < n; ++i) {
        ++ret[i][s[i] - 'a'];
        ret[i + 1] = ret[i];
      }
      return ret;
    };
    auto a = get_prefs(s);
    auto b = get_prefs(t);
    {
      string a = s;
      string b = t;
      sort((a).begin(), (a).end());
      sort((b).begin(), (b).end());
      if (a != b) {
        cout << "-1" << endl;
        continue;
      }
    }
    auto f = vector<vector<int>>(n + 1, vector<int>(n + 1, -1));
    f[0][0] = 0;
    for (int i = 0; i <= n; ++i) {
      for (int j = 0; j <= n; ++j) {
        if (f[i][j] == -1) {
          continue;
        }
        if (i < n) {
          chmax(f[i + 1][j], f[i][j]);
        }
        if (j < n) {
          chmax(f[i][j + 1], f[i][j]);
        }
        if (i < n && j < n && s[i] == t[j]) {
          bool ok = true;
          for (int x = 0; x < 26; ++x) {
            ok &= a[i][x] <= b[j][x];
          }
          if (ok) {
            chmax(f[i + 1][j + 1], f[i][j] + 1);
          }
        }
      }
    }
    cout << n - f[n][n] << endl;
  }
  cerr << "Time execute: " << clock() / (double)CLOCKS_PER_SEC << " sec"
       << endl;
  return 0;
}
