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;
ifstream fin("AAtest.in.txt");
int n, ma[2005], t, a, uu;
vector<int> v['z' + 1], v1['z' + 1];
string s1, s2;
bool b;
int esi(int o, char c) {
  int l = 0, r = v[c].size(), k;
  while (l < r) {
    k = (l + r) / 2;
    if (v[c][k] >= o)
      r = k;
    else
      l = k + 1;
  }
  if (l == v[c].size()) return -1;
  return v[c][r];
}
int mit(char c, int o) {
  int l = 0, r = v1[c].size() - 1, k;
  while (l < r) {
    k = (l + r) / 2;
    if (v1[c][k] == o)
      r = k, l = k;
    else if (v1[c][k] > o)
      r = k - 1;
    else
      l = k + 1;
  }
  return l;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cerr.tie(0);
  cin >> t;
  while (t--) {
    cin >> n >> s1 >> s2;
    for (char c = 'a'; c <= 'z'; c++) v[c].resize(0), v1[c].resize(0);
    memset(ma, -1, sizeof(ma));
    for (int i = 0; i < n; i++) v[s2[i]].push_back(i), v1[s1[i]].push_back(i);
    b = 0;
    uu = 0;
    for (char c = 'a'; c <= 'z'; c++)
      if (v[c].size() != v1[c].size()) {
        b = 1;
        break;
      }
    if (b) {
      cout << -1 << "\n";
      continue;
    }
    ma[1] = v[s1[0]][0];
    for (int i = 1; i < n; i++) {
      for (int j = i; j >= 1; j--)
        if (ma[j] != -1) {
          a = esi(max(uu, max(i, ma[j] + 1)), s1[i]);
          if (a != -1)
            if (ma[j + 1] == -1 || a < ma[j + 1]) ma[j + 1] = a;
        }
      uu = max(uu, v[s1[i]][mit(s1[i], i)] + 1);
    }
    for (int i = n; i > 0; i--)
      if (ma[i] != -1) {
        cout << n - i << "\n";
        break;
      }
  }
}
