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;
using ll = long long;
using ld = long double;
using ull = unsigned long long;
template <typename T1, typename T2>
inline void chkmin(T1 &x, const T2 &y) {
  if (x > y) x = y;
}
template <typename T1, typename T2>
inline void chkmax(T1 &x, const T2 &y) {
  if (x < y) x = y;
}
const int INF = 1e9 + 228 + 1337;
bool check(string s, string t) {
  sort(s.begin(), s.end());
  sort(t.begin(), t.end());
  return s == t;
}
void solve() {
  int n;
  vector<string> a(2);
  cin >> n >> a[0] >> a[1];
  if (!check(a[0], a[1])) {
    cout << -1 << "\n";
    return;
  }
  vector<vector<int>> dp(n + 1, vector<int>(n + 1, INF));
  dp[0][0] = 0;
  vector<vector<vector<int>>> cnt(
      2, vector<vector<int>>(n + 1, vector<int>(26, 0)));
  reverse(a[0].begin(), a[0].end());
  reverse(a[1].begin(), a[1].end());
  for (int it = 0; it < 2; it++) {
    for (int i = 1; i <= n; i++) {
      for (int alpha = 0; alpha < 26; alpha++) {
        cnt[it][i][alpha] = cnt[it][i - 1][alpha];
      }
      cnt[it][i][a[it][i - 1] - 'a']++;
    }
  }
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= i; j++) {
      if (i + 1 <= n && a[0][i] == a[1][j]) {
        assert(j + 1 <= n);
        chkmin(dp[i + 1][j + 1], dp[i][j]);
      }
      char x = a[1][j] - 'a';
      if (j + 1 <= n && cnt[0][i][x] > cnt[1][j][x]) {
        chkmin(dp[i][j + 1], dp[i][j]);
      }
      if (i + 1 <= n) {
        chkmin(dp[i + 1][j], dp[i][j] + 1);
      }
    }
  }
  int ans = -1;
  if (dp[n][n] != INF) {
    ans = dp[n][n];
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
