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;
int const nax = 2048;
int const inf = 2e8;
int n;
int dp[nax][nax];
int sum_s[26][nax];
int sum_t[26][nax];
char s[nax];
char t[nax];
int rec(int x, int y) {
  if (x > y || x < 0 || y < 0) return inf;
  if (y == 0) return 0;
  if (dp[x][y] != -1) return dp[x][y];
  int &res = dp[x][y];
  res = inf;
  if (x > 0 && s[x] == t[y]) {
    res = min(res, rec(x - 1, y - 1));
  }
  if (x > 0) res = min(res, rec(x - 1, y) + 1);
  int c = t[y] - 'a';
  if (sum_t[c][y] > sum_s[c][x]) {
    res = min(res, rec(x, y - 1));
  }
  return res;
}
int solve() {
  scanf("%d%s%s", &n, s + 1, t + 1);
  for (int i = 0; i <= n; ++i)
    for (int j = 0; j <= n; ++j) {
      dp[i][j] = -1;
    }
  for (int i = 0; i < 26; ++i) {
    sum_s[i][n + 1] = sum_t[i][n + 1] = 0;
  }
  for (int i = 1; i <= n; ++i)
    for (int c = 0; c < 26; ++c) {
      sum_s[c][i] = sum_s[c][i - 1];
      sum_t[c][i] = sum_t[c][i - 1];
      sum_s[s[i] - 'a'][i]++;
      sum_t[t[i] - 'a'][i]++;
    }
  for (int i = 0; i < 26; ++i) {
    sum_s[i][n + 1] = sum_s[i][n];
    sum_t[i][n + 1] = sum_t[i][n];
  }
  for (int c = 0; c < 26; ++c) {
    if (sum_s[c][n] != sum_t[c][n]) return -1;
  }
  return rec(n, n);
}
int main() {
  int T;
  for (cin >> T; T--;) {
    printf("%d\n", solve());
  }
}
