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 n, a[2][2069], ps[2][26][2069], dp[2069][2069], inf = 1e18;
int main() {
  long long t, rr, i, j, r, ii, z;
  char ch;
  scanf("%lld", &t);
  for (rr = 0; rr < t; rr++) {
    scanf("%lld", &n);
    for (ii = 0; ii < 2; ii++) {
      for (i = 1; i <= n; i++) {
        scanf(" %c", &ch);
        a[ii][i] = ch - 'a';
        for (j = 0; j < 26; j++) {
          ps[ii][j][i] = ps[ii][j][i - 1];
        }
        ps[ii][a[ii][i]][i]++;
      }
    }
    dp[0][0] = 0;
    for (i = 0; i <= n; i++) {
      for (j = !i; j <= n; j++) {
        dp[i][j] = -inf;
        for (r = 0; r < 26; r++) {
          if (ps[0][r][i] > ps[1][r][j]) {
            r = -1;
            break;
          }
        }
        if (r != -1) {
          if (i) {
            dp[i][j] = max(dp[i][j], dp[i - 1][j]);
          }
          if (j) {
            dp[i][j] = max(dp[i][j], dp[i][j - 1]);
          }
          if (i && j && a[0][i] == a[1][j]) {
            dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + 1);
          }
        }
      }
    }
    if (dp[n][n] != -inf) {
      z = n - dp[n][n];
    } else {
      z = -1;
    }
    printf("%lld\n", z);
  }
}
