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 long long N = 400005;
const long long INF = 0x3f3f3f3f3f3f3f3f;
long long n, T, dp[2005][2005];
long long prs[30][2005], prt[30][2005];
string s, t;
long long init() {
  for (long long i = 1; i <= n; i++)
    for (long long j = i; j <= n; j++) dp[i][j] = INF;
  for (long long i = 0; i < 26; i++) prs[0][i] = prt[0][i] = 0;
  for (long long i = 1; i <= n; i++) {
    for (long long j = 0; j < 26; j++) {
      prs[j][i] = prs[j][i - 1];
      prt[j][i] = prt[j][i - 1];
    }
    prs[s[i] - 'a'][i]++;
    prt[t[i] - 'a'][i]++;
  }
  for (long long i = 0; i < 26; i++)
    if (prs[i][n] != prt[i][n]) return 1;
  return 0;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> T;
  while (T--) {
    cin >> n;
    cin >> s;
    cin >> t;
    s = ' ' + s;
    t = ' ' + t;
    if (init()) {
      cout << -1 << '\n';
      continue;
    }
    for (long long i = 1; i <= n; i++) {
      for (long long j = i; j <= n; j++) {
        dp[i][j] = dp[i - 1][j] + 1;
        if (s[i] == t[j]) dp[i][j] = min(dp[i][j], dp[i - 1][j - 1]);
        if (prs[t[j] - 'a'][i] < prt[t[j] - 'a'][j])
          dp[i][j] = min(dp[i][j], dp[i][j - 1]);
      }
    }
    cout << dp[n][n] << '\n';
  }
  return 0;
}
