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;
long long mod = 1000000007;
int cache[2005][2005];
char s1[2005], s2[2005];
int suf1[2005][30], suf2[2005][30];
int dfs(int l, int r) {
  if (r == 0) return 0;
  int &ans = cache[l][r];
  if (ans != -1) return ans;
  char c2 = s2[r] - 'a', c1 = s1[l] - 'a';
  ans = 0x3f3f3f3f;
  if (suf1[l + 1][c2] > suf2[r + 1][c2]) ans = min(ans, dfs(l, r - 1));
  if (l) {
    ans = min(ans, dfs(l - 1, r) + 1);
    if (c1 == c2) ans = min(ans, dfs(l - 1, r - 1));
  }
  return ans;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    scanf("%s%s", s1 + 1, s2 + 1);
    for (int i = 0; i <= n + 1; i++)
      for (int j = 0; j <= n + 1; j++) cache[i][j] = -1;
    for (int i = 0; i <= 'z' - 'a'; i++) suf1[n + 1][i] = suf2[n + 1][i] = 0;
    for (int i = n; i >= 1; i--) {
      for (int j = 0; j <= 'z' - 'a'; j++) {
        suf1[i][j] = suf1[i + 1][j];
        suf2[i][j] = suf2[i + 1][j];
      }
      suf1[i][s1[i] - 'a']++;
      suf2[i][s2[i] - 'a']++;
    }
    int ok = 1;
    for (int i = 0; i <= 'z' - 'a'; i++)
      if (suf1[1][i] != suf2[1][i]) ok = 0;
    if (!ok)
      printf("-1\n");
    else {
      int ans = dfs(n, n);
      printf("%d\n", ans);
    }
  }
  return 0;
}
