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 BIG_INF = 4e18;
const long long INF = 2e9;
const long long mod = 1e9 + 7;
long long fast_exp(long long a, long long b) {
  if (b <= 0)
    return 1;
  else {
    long long res = 1;
    res = fast_exp(a, b / 2);
    res = res % mod;
    res = (res * res) % mod;
    if (b % 2 == 1) res = (res * a) % mod;
    return res;
  }
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long true_rand(long long n) {
  uniform_int_distribution<long long> uid(0, n - 1);
  return uid(rng);
}
const long long N = 5e5 + 1000;
const long double pi = acos(-1.0);
using cd = complex<double>;
const double PI = acos(-1);
double eps = 1e-12;
string s, t;
int dp[2005][2005] = {};
int done[2005][2005] = {};
int frqs[2005][26] = {};
int frqt[2005][26] = {};
int moves_needed(int ptrs, int ptrt) {
  if (ptrs == -1 || ptrt == -1) {
    return abs(max(ptrs, ptrt) + 1);
  }
  if (done[ptrs][ptrt] == 1) return dp[ptrs][ptrt];
  int ans = INF;
  if (s[ptrs] == t[ptrt]) {
    ans = min(ans, moves_needed(ptrs - 1, ptrt - 1));
  } else {
    if (frqs[ptrs + 1][t[ptrt] - 'a'] - frqt[ptrt + 1][t[ptrt] - 'a'] > 0)
      ans = min(ans, moves_needed(ptrs, ptrt - 1) + 1);
    ans = min(ans, moves_needed(ptrs - 1, ptrt));
  }
  done[ptrs][ptrt] = 1;
  dp[ptrs][ptrt] = ans;
  return dp[ptrs][ptrt];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int tc;
  cin >> tc;
  while (tc--) {
    int n;
    cin >> n;
    for (int i = 0; i < n + 4; i++) {
      for (int j = 0; j < n + 4; j++) {
        dp[i][j] = INF;
        done[i][j] = 0;
      }
      for (int j = 0; j < 26; j++) {
        frqs[i][j] = 0;
        frqt[i][j] = 0;
      }
    }
    cin >> s;
    cin >> t;
    for (int i = n - 1; i >= 0; i--) {
      for (int ch = 0; ch < 26; ch++) {
        int add = 0;
        if (s[i] - 'a' == ch) add = 1;
        frqs[i][ch] = frqs[i + 1][ch] + add;
      }
      for (int ch = 0; ch < 26; ch++) {
        int add = 0;
        if (t[i] - 'a' == ch) add = 1;
        frqt[i][ch] = frqt[i + 1][ch] + add;
      }
    }
    int chk = 1;
    for (int i = 0; i < 26; i++) {
      if (frqs[0][i] != frqt[0][i]) chk = 0;
    }
    if (chk == 1)
      cout << moves_needed(n - 1, n - 1) << '\n';
    else
      cout << -1 << endl;
  }
  return 0;
}
