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;
using int32 = int_fast32_t;
using int64 = int_fast64_t;
const int32 INF = 1e9;
const int32 MOD = 1e9 + 7;
const int64 LLINF = 1e18;
void solve() {
  int32 n;
  cin >> n;
  string s, t;
  cin >> s >> t;
  vector<vector<int32>> cnts(n + 1, vector<int32>(26, 0)),
      cntt(n + 1, vector<int32>(26, 0));
  for (int64 i = (n - 1); i >= (0); i--) {
    for (int64 j = 0; j < (26); ++j) {
      cnts[i][j] = cnts[i + 1][j];
      cntt[i][j] = cntt[i + 1][j];
    }
    cnts[i][s[i] - 'a']++;
    cntt[i][t[i] - 'a']++;
  }
  for (int64 j = 0; j < (26); ++j) {
    if (cnts[0][j] != cntt[0][j]) {
      cout << (-1) << "\n";
      return;
    }
  }
  vector<vector<int32>> dp(n + 1, vector<int32>(n + 1, INF));
  for (int64 j = 0; j < (n + 1); ++j) dp[0][j] = 0;
  for (int64 i = 0; i < (n); ++i)
    for (int64 j = 0; j < (n); ++j) {
      if (i > j) continue;
      dp[i + 1][j + 1] = min(dp[i + 1][j + 1], dp[i][j + 1] + 1);
      if (s[i] == t[j]) dp[i + 1][j + 1] = min(dp[i + 1][j + 1], dp[i][j]);
      if (cnts[i + 1][t[j] - 'a'] > cntt[j + 1][t[j] - 'a'])
        dp[i + 1][j + 1] = min(dp[i + 1][j + 1], dp[i + 1][j]);
    }
  cout << (dp[n][n]) << "\n";
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int32 t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
