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 int MAXN = 2000 + 2;
int suf_a[MAXN][26];
int suf_b[MAXN][26];
int f[MAXN][MAXN];
int main() {
  int testcases;
  cin >> testcases;
  memset(suf_a, 0, sizeof(suf_a));
  memset(suf_b, 0, sizeof(suf_b));
  memset(f, 0, sizeof(f));
  while (testcases--) {
    int n;
    string a, b;
    cin >> n >> a >> b;
    a = "#" + a;
    b = "$" + b;
    for (int i = n; i >= 1; --i) {
      for (int j = 0; j < 26; ++j) {
        suf_a[i][j] = suf_a[i + 1][j];
        suf_b[i][j] = suf_b[i + 1][j];
      }
      char ca = a[i] - 'a';
      char cb = b[i] - 'a';
      suf_a[i][ca]++;
      suf_b[i][cb]++;
    }
    bool flag = true;
    for (int i = 0; i < 26 && flag; ++i)
      if (suf_a[1][i] != suf_b[1][i]) flag = false;
    if (!flag) {
      cout << -1 << endl;
      continue;
    }
    for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= n; ++j) {
        f[i][j] = max(f[i - 1][j], f[i][j - 1]);
        if (a[i] == b[j]) {
          bool flag = true;
          for (int k = 0; k < 26 && flag; ++k)
            if (suf_a[i + 1][k] < suf_b[j + 1][k]) {
              flag = false;
            }
          if (flag) f[i][j] = max(f[i - 1][j - 1] + 1, f[i][j]);
        }
      }
    cout << n - f[n][n] << endl;
  }
  return 0;
}
