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 f[509], g[509], l[2009];
int dp[2009][2009];
void solve() {
  int n;
  string s, t, cs, ct;
  cin >> n >> s >> t;
  cs = s, ct = t;
  sort(cs.begin(), cs.end());
  sort(ct.begin(), ct.end());
  if (cs != ct) {
    cout << -1 << endl;
    return;
  }
  for (int c = 'a'; c <= 'z'; c++) f[c] = 0;
  s = '#' + s;
  t = '#' + t;
  for (int i = 1; i <= n; i++) {
    for (int c = 'a'; c <= 'z'; c++) g[c] = 0;
    f[t[i]]++;
    int k = 1;
    while (k <= n) {
      int x = s[k];
      g[x]++;
      if (g[x] > f[x])
        break;
      else
        k++;
    }
    l[i] = k - 1;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      dp[i][j] = max(dp[i][j - 1], dp[i - 1][j]);
      if (s[i] == t[j] && i <= l[j])
        dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + 1);
    }
  }
  cout << n - dp[n][n] << endl;
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
}
