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 dp[2005][2005];
int cnt1[2005][26], cnt2[2005][26];
char s[2005], x[2005];
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    memset(cnt1, 0, sizeof(cnt1));
    memset(cnt2, 0, sizeof(cnt2));
    int n;
    scanf("%d", &n);
    scanf("%s%s", s + 1, x + 1);
    for (int i = 1; i <= n; i++) {
      cnt1[i][s[i] - 'a']++;
      cnt2[i][x[i] - 'a']++;
      for (int j = 0; j < 26; j++)
        cnt1[i][j] += cnt1[i - 1][j], cnt2[i][j] += cnt2[i - 1][j];
    }
    for (int i = 1; i <= n + 1; i++)
      for (int j = 1; j <= n + 1; j++) dp[i][j] = 1e9;
    dp[1][1] = 0;
    for (int i = 1; i <= n + 1; i++) {
      for (int j = i; j <= n + 1; j++) {
        if (i <= n && j <= n)
          if (s[i] == x[j]) dp[i + 1][j + 1] = min(dp[i + 1][j + 1], dp[i][j]);
        if (i <= n && j <= n)
          if (cnt1[n][x[j] - 'a'] - cnt1[i][x[j] - 'a'] > 0)
            dp[i][j + 1] = min(dp[i][j + 1], dp[i][j] + 1);
        if (cnt2[j - 1][s[i] - 'a'] - cnt1[i - 1][s[i] - 'a'] > 0)
          dp[i + 1][j] = min(dp[i + 1][j], dp[i][j]);
      }
    }
    if (dp[n + 1][n + 1] == 1e9) dp[n + 1][n + 1] = -1;
    printf("%d\n", dp[n + 1][n + 1]);
  }
}
