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 f = 1, ans = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ans = ans * 10 + c - '0';
    c = getchar();
  }
  return f * ans;
}
const int MAXN = 2011;
int f[MAXN][MAXN], Num[2][MAXN][27], N, cas;
char S[MAXN], T[MAXN];
bool Get(int x, int y) {
  for (int i = 0; i < 26; i++)
    if (Num[0][x][i] < Num[1][y][i]) return 0;
  return 1;
}
int main() {
  cas = read();
  while (cas--) {
    N = read();
    scanf("%s%s", S + 1, T + 1);
    for (int i = 0; i <= N + 1; i++)
      for (int j = 0; j <= N + 1; j++) f[i][j] = INT_MIN;
    for (int i = 0; i <= N + 1; i++)
      for (int j = 0; j < 26; j++) Num[0][i][j] = Num[1][i][j] = 0;
    for (int i = N; i >= 1; i--) {
      for (int j = 0; j < 26; j++) Num[0][i][j] = Num[0][i + 1][j];
      Num[0][i][S[i] - 'a']++;
      for (int j = 0; j < 26; j++) Num[1][i][j] = Num[1][i + 1][j];
      Num[1][i][T[i] - 'a']++;
    }
    bool ff = 1;
    for (int i = 0; i < 26; i++) ff &= (Num[0][1][i] == Num[1][1][i]);
    if (!ff) {
      printf("-1\n");
      continue;
    }
    f[0][0] = 0;
    for (int i = 0; i <= N; i++) f[0][i] = f[i][0] = 0;
    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 (Get(i, j) && S[i] == T[j])
          f[i][j] = max(f[i][j], f[i - 1][j - 1] + 1);
      }
    }
    printf("%d\n", N - f[N][N]);
  }
  return 0;
}
