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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long power(long long a, long long n) {
  a %= 1000000007;
  if (n == 1) return a;
  if (n == 0) return 1;
  if (n % 2)
    return (a * (power((a * a) % 1000000007, n / 2) % 1000000007)) % 1000000007;
  return power((a * a) % 1000000007, n / 2) % 1000000007;
}
const long long inf = (long long)1e18;
long long inverse(long long x) { return power(x, 1000000007 - 2) % 1000000007; }
long long dp[2005][2005];
void solve() {
  long long n;
  cin >> n;
  string s, t;
  cin >> s >> t;
  s = '#' + s;
  t = '#' + t;
  long long cnt1[30][n + 5], cnt2[30][n + 5];
  memset(cnt1, 0, sizeof cnt1);
  memset(cnt2, 0, sizeof cnt2);
  for (long long i = 1; i < n + 1; ++i) {
    for (long long j = 0; j < 26; ++j) cnt1[j][i] = cnt1[j][i - 1];
    cnt1[s[i] - 'a'][i]++;
  }
  for (long long i = 1; i < n + 1; ++i) {
    for (long long j = 0; j < 26; ++j) cnt2[j][i] = cnt2[j][i - 1];
    cnt2[t[i] - 'a'][i]++;
  }
  for (long long i = 0; i < 30; ++i)
    if (cnt1[i][n] != cnt2[i][n]) {
      cout << "-1\n";
      return;
    }
  for (long long i = 1; i < n + 1; ++i) {
    for (long long j = 1; j < n + 1; ++j) dp[i][j] = inf;
  }
  for (long long i = 1; i < n + 1; ++i) {
    for (long long j = 1; j < n + 1; ++j) {
      if (s[i] == t[j]) dp[i][j] = dp[i - 1][j - 1];
      if (cnt1[t[j] - 'a'][n] - cnt1[t[j] - 'a'][i] >
          cnt2[t[j] - 'a'][n] - cnt2[t[j] - 'a'][j]) {
        dp[i][j] = min(dp[i][j], dp[i][j - 1]);
      }
      dp[i][j] = min(dp[i][j], 1 + dp[i - 1][j]);
    }
  }
  cout << dp[n][n] << "\n";
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long T = 1;
  cin >> T;
  for (long long t = 1; t < T + 1; ++t) {
    solve();
  }
  return 0;
}
