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 N = 2020;
const int INF = 0x3f3f3f3f;
int n;
char s[N];
char t[N];
int f[N][N];
int suf1[N][26];
int suf2[N][26];
int dp(int i, int j) {
  if (j == 0) return 0;
  int &res = f[i][j];
  if (res != -1) return res;
  res = INF;
  if (i != 0) {
    res = 1 + dp(i - 1, j);
    if (s[i] == t[j]) res = min(res, dp(i - 1, j - 1));
  }
  int ch = t[j] - 'a';
  if (suf1[i + 1][ch] > suf2[j + 1][ch]) res = min(res, dp(i, j - 1));
  return res;
}
int solve() {
  scanf("%d", &n);
  scanf("%s", s + 1);
  scanf("%s", t + 1);
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= n; j++) f[i][j] = -1;
  for (int i = 1; i <= n + 1; i++)
    for (int j = 0; j < 26; j++) suf2[i][j] = suf1[i][j] = 0;
  for (int i = n; i >= 1; i--) {
    suf1[i][s[i] - 'a']++;
    suf2[i][t[i] - 'a']++;
    for (int j = 0; j < 26; j++) {
      suf1[i][j] += suf1[i + 1][j];
      suf2[i][j] += suf2[i + 1][j];
    }
  }
  for (int i = 0; i < 26; i++)
    if (suf1[1][i] != suf2[1][i]) return -1;
  return dp(n, n);
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) printf("%d\n", solve());
  return 0;
}
