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 T>
inline T read() {
  T x = 0, w = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') w = -w;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  return x * w;
}
template <class T>
inline T read(T& x) {
  return x = read<T>();
}
const int N = 2e3 + 10, inf = 1e9;
char s[N], t[N];
int suf[N][26], suf2[N][26];
int dp[N][N];
void real_main() {
  int n = read<int>();
  scanf("%s%s", s + 1, t + 1);
  fill(suf[n + 1], suf[n + 1] + 26, 0), fill(suf2[n + 1], suf2[n + 1] + 26, 0);
  for (int i = n; i >= 1; --i) {
    copy(suf[i + 1], suf[i + 1] + 26, suf[i]),
        copy(suf2[i + 1], suf2[i + 1] + 26, suf2[i]);
    ++suf[i][s[i] - 'a'], ++suf2[i][t[i] - 'a'];
  }
  for (int i = 0; i <= n; ++i) dp[i][0] = 0;
  for (int j = 1; j <= n; ++j)
    for (int i = 0; i <= n; ++i) {
      int& ans = dp[i][j] = inf;
      if (i > 0) {
        ans = min(ans, 1 + dp[i - 1][j]);
        if (s[i] == t[j]) ans = min(ans, dp[i - 1][j - 1]);
      }
      int c = t[j] - 'a';
      if (suf[i + 1][c] > suf2[j + 1][c]) ans = min(ans, dp[i][j - 1]);
    }
  printf("%d\n", dp[n][n] == inf ? -1 : dp[n][n]);
}
int main() {
  for (int T = read<int>(); T--;) real_main();
  return 0;
}
