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 <class T1, class T2, class T3 = hash<T1>>
using umap = unordered_map<T1, T2, T3>;
template <class T>
using uset = unordered_set<T>;
template <class T>
using vec = vector<T>;
const long long infll = numeric_limits<long long>::max() >> 1;
const int inf = numeric_limits<int>::max() >> 1;
const int N = 2005;
const int L = 26;
int n, tc;
int dp[N][N];
int cnt[N][L];
int cns[N][L];
string t;
string s;
void input() {
  cin >> n;
  cin >> s;
  cin >> t;
  s = '#' + s;
  t = '#' + t;
  for (int i = 0; i < L; ++i) {
    cnt[n + 1][i] = 0;
    cns[n + 1][i] = 0;
  }
  for (int i = n; i >= 1; --i) {
    for (int j = 0; j < L; ++j) {
      cnt[i][j] = cnt[i + 1][j];
    }
    cnt[i][t[i] - 'a']++;
  }
  for (int i = n; i >= 1; --i) {
    for (int j = 0; j < L; ++j) {
      cns[i][j] = cns[i + 1][j];
    }
    cns[i][s[i] - 'a']++;
  }
}
string sorted(string str) {
  sort((str).begin(), (str).end());
  return str;
}
int calc(int i, int j) {
  if (i == -1 && j == -1) {
    return 0;
  }
  if (dp[i][j] != -1) {
    return dp[i][j];
  }
  if (t[i] == s[j]) {
    return dp[i][j] = calc(i - 1, j - 1);
  }
  int res = inf;
  int c = cns[j + 1][t[i] - 'a'] - cnt[i + 1][t[i] - 'a'];
  if (c >= 1) {
    res = min(res, calc(i - 1, j));
  }
  if (j) {
    res = min(res, calc(i, j - 1) + 1);
  }
  return dp[i][j] = res;
}
void solve() {
  if (sorted(s) != sorted(t)) {
    cout << "-1\n";
    return;
  }
  memset(dp, -1, sizeof dp);
  cout << calc(n, n) << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> tc;
  while (tc--) {
    input();
    solve();
  }
}
