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];
bool pos[2005][2005];
int cnt[26];
int main() {
  ios::sync_with_stdio(0);
  int _T;
  cin >> _T;
  for (int(_t) = 0; (_t) < (_T); ++(_t)) {
    for (int(i) = 0; (i) < (26); ++(i)) cnt[i] = 0;
    int n;
    cin >> n;
    string s, t;
    cin >> s >> t;
    for (char c : s) ++cnt[c - 'a'];
    for (char c : t) --cnt[c - 'a'];
    bool ok = true;
    for (int(i) = 0; (i) < (26); ++(i))
      if (cnt[i] != 0) ok = false;
    if (!ok) {
      cout << -1 << "\n";
      continue;
    }
    for (int(i) = 0; (i) < (26); ++(i)) cnt[i] = 0;
    int nonz = 0, currNonz;
    int currCnt[26];
    for (int(i) = 0; (i) < (n + 1); ++(i)) {
      if (i > 0 && cnt[s[i - 1] - 'a']++ == 0) ++nonz;
      currNonz = nonz;
      for (int(j) = 0; (j) < (26); ++(j)) currCnt[j] = cnt[j];
      for (int(j) = 0; (j) < (n + 1); ++(j)) {
        if (j > 0 && --currCnt[t[j - 1] - 'a'] == 0) --currNonz;
        pos[i][j] = (currNonz == 0);
      }
    }
    for (int(i) = 0; (i) < (n + 1); ++(i))
      for (int j = i; j <= n; ++j) {
        dp[i][j] = 1000000;
        if (i == 0) {
          dp[i][j] = 0;
          continue;
        }
        if (!pos[i][j]) continue;
        if (s[i - 1] == t[j - 1]) dp[i][j] = dp[i - 1][j - 1];
        dp[i][j] = min(dp[i][j], dp[i - 1][j] + 1);
        if (pos[i][j - 1]) dp[i][j] = min(dp[i][j], dp[i][j - 1]);
      }
    cout << dp[n][n] << "\n";
  }
  return 0;
}
