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 mod1 = 998244353, mod = 1e9 + 7;
const int MAXN = 100005, MAXM = 200010;
int dp[2020][2020];
int cntS[2020][26], cntT[2020][26];
void Solve() {
  string s, t;
  int n;
  cin >> n >> s >> t;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 26; j++) {
      cntS[i + 1][j] = cntS[i][j] + (s[i] == j + 'a');
      cntT[i + 1][j] = cntT[i][j] + (t[i] == j + 'a');
    }
  }
  for (int i = 0; i < 26; i++) {
    if (cntS[n][i] != cntT[n][i]) {
      cout << -1 << endl;
      return;
    }
  }
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= n; j++) {
      dp[i][j] = -1;
    }
  }
  dp[0][0] = 0;
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= n; j++) {
      dp[i + 1][j] = max(dp[i + 1][j], dp[i][j]);
      dp[i][j + 1] = max(dp[i][j + 1], dp[i][j]);
      if (s[i] == t[j]) {
        bool check = true;
        for (int k = 0; k < 26; k++) {
          check &= (cntS[i][k] <= cntT[j][k]);
        }
        if (check) {
          dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j] + 1);
        }
      }
    }
  }
  cout << n - dp[n][n] << endl;
}
signed main() {
  if (fopen(""
            ".inp",
            "r")) {
    freopen(
        ""
        ".inp",
        "r", stdin);
    freopen(
        ""
        ".out",
        "w", stdout);
  };
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int TC = 1;
  cin >> TC;
  while (TC--) Solve();
  return 0;
}
