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 maxn = 2005;
const int inf = 2147483647;
const double eps = 1e-6;
const long long INF = 9223372036854775807ll;
int t, n, dp[maxn][maxn], cnt[2][27][maxn];
char ss[maxn], tt[maxn];
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d%s%s", &n, ss + 1, tt + 1);
    for (register int i = 0; i < (27); ++i)
      for (register int j = 1; j < (n + 1); ++j) {
        cnt[0][i][j] = cnt[0][i][j - 1] + (ss[j] - 'a' == i);
        cnt[1][i][j] = cnt[1][i][j - 1] + (tt[j] - 'a' == i);
      }
    for (register int i = 0; i < (n + 1); ++i)
      for (register int j = 0; j < (n + 1); ++j) dp[i][j] = inf / 2;
    dp[n][n] = 0;
    for (register int i = n; i >= (0); --i)
      for (register int j = n; j >= (i); --j) {
        if (i != n && j != n && ss[i + 1] == tt[j + 1])
          dp[i][j] = min(dp[i][j], dp[i + 1][j + 1]);
        if (j != n &&
            cnt[0][tt[j + 1] - 'a'][n] - cnt[0][tt[j + 1] - 'a'][i] >
                cnt[1][tt[j + 1] - 'a'][n] - cnt[1][tt[j + 1] - 'a'][j + 1])
          dp[i][j] = min(dp[i][j], dp[i][j + 1]);
        if (i != n) dp[i][j] = min(dp[i][j], dp[i + 1][j] + 1);
      }
    if (dp[0][0] < inf / 2)
      printf("%d\n", dp[0][0]);
    else
      printf("-1\n");
  }
}
