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 = 2e3 + 3;
const int ALF = 26;
int pref[MAXN][ALF];
int pref2[MAXN][ALF];
int dp[MAXN][MAXN];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int z, n, x;
  bool por;
  string s1, s2;
  cin >> z;
  while (z--) {
    cin >> n;
    cin >> s1 >> s2;
    s1 = '#' + s1;
    s2 = '#' + s2;
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j < ALF; j++) {
        pref[i][j] = pref[i - 1][j];
        pref2[i][j] = pref2[i - 1][j];
      }
      pref[i][s1[i] - 'a']++;
      pref2[i][s2[i] - 'a']++;
    }
    por = false;
    for (int i = 0; i < ALF; i++) {
      if (pref[n][i] != pref2[n][i]) por = true;
    }
    if (por) {
      cout << "-1\n";
      continue;
    }
    for (int i = 1; i <= n; i++) {
      for (int j = i; j <= n; j++) {
        dp[i][j] = dp[i - 1][j] + 1;
        if (s1[i] == s2[j]) dp[i][j] = min(dp[i][j], dp[i - 1][j - 1]);
        x = s2[j] - 'a';
        if (pref[n][x] - pref[i][x] > pref2[n][x] - pref2[j][x])
          dp[i][j] = min(dp[i][j], dp[i][j - 1]);
      }
    }
    cout << dp[n][n] << '\n';
  }
  return 0;
}
