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 long long int mod = 998244353;
const long double error = 2e-6;
const long double PI = acosl(-1);
mt19937 rng((unsigned)chrono::system_clock::now().time_since_epoch().count());
inline long long int MOD(long long int x, long long int m = mod) {
  long long int y = x % m;
  return (y >= 0) ? y : y + m;
}
const int inf = 1e9;
const long long int infl = 1061109567;
const int nmax = 2000 + 10;
char first[nmax], second[nmax];
int memo[nmax][nmax];
int cnt[128];
bool matchable[nmax][nmax];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int tc;
  cin >> tc;
  for (int cs = 1; cs <= tc; cs++) {
    int len;
    cin >> len;
    cin >> first;
    cin >> second;
    for (int i = 0; i < len; i++) cnt[first[i]]++;
    for (int i = 0; i < len; i++) cnt[second[i]]--;
    vector<int> seencnt1(128);
    for (int i = len - 1; i >= 0; i--) {
      seencnt1[first[i]]++;
      vector<int> seencnt2(128);
      bool problem = false;
      for (int j = len - 1; j >= 0; j--) {
        seencnt2[second[j]]++;
        if (seencnt2[second[j]] > seencnt1[second[j]]) problem = true;
        if (!problem && first[i] == second[j])
          matchable[i][j] = true;
        else
          matchable[i][j] = false;
      }
    }
    bool okay = true;
    for (int i = 0; i < 128; i++) {
      okay &= (cnt[i] == 0);
      cnt[i] = 0;
    }
    if (!okay) {
      cout << -1 << "\n";
      continue;
    }
    for (int i = 0; i <= len; i++) {
      for (int j = 0; j <= len; j++) {
        if (i == 0 || j == 0) {
          memo[i][j] = 0;
          continue;
        }
        memo[i][j] = max(memo[i - 1][j], memo[i][j - 1]);
        if (matchable[i - 1][j - 1]) {
          memo[i][j] = max(memo[i][j], memo[i - 1][j - 1] + 1);
        }
      }
    }
    int ans = len - memo[len][len];
    cout << ans << "\n";
  }
  return 0;
}
