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;
inline int read() {
  int ans = 0, f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) ans = (ans << 3) + (ans << 1) + (c ^ 48);
  return ans * f;
}
inline int cmin(int &a, int b) { return a = min(a, b); }
inline int cmax(int &a, int b) { return a = max(a, b); }
int n;
int f[2005][2005];
char s[2005], t[2005];
int sufs[28][2005], suft[28][2005];
int main() {
  int T = read();
  while (T--) {
    n = read();
    for (int i = 0; i <= n; i++)
      for (int j = 0; j <= n; j++) f[i][j] = 1e9;
    cin >> s + 1 >> t + 1;
    memset(sufs, 0, sizeof(sufs));
    memset(suft, 0, sizeof(suft));
    for (int i = n; i >= 1; i--) {
      for (int j = 0; j < 26; j++)
        sufs[j][i] = sufs[j][i + 1], suft[j][i] = suft[j][i + 1];
      sufs[s[i] - 'a'][i]++;
      suft[t[i] - 'a'][i]++;
    }
    int flag = 0;
    for (int i = 0; i < 26; i++)
      if (sufs[i][1] != suft[i][1]) flag = 1;
    if (flag) {
      cout << -1 << endl;
      continue;
    }
    for (int i = 0; i <= n; i++) f[0][i] = 0;
    for (int i = 0; i <= n; i++) {
      for (int j = i; j <= n; j++) {
        if (!j) break;
        if (i) {
          f[i][j] = 1 + f[i - 1][j];
          if (s[i] == t[j]) f[i][j] = min(f[i][j], f[i - 1][j - 1]);
        }
        if (sufs[t[j] - 'a'][i + 1] > suft[t[j] - 'a'][j + 1])
          f[i][j] = min(f[i][j], f[i][j - 1]);
      }
    }
    cout << f[n][n] << endl;
  }
  return 0;
}
