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>
#pragma GCC optimize(3)
#pragma GCC optimize(2)
using namespace std;
template <typename T>
inline T read() {
  T s = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    s = (s << 3) + (s << 1) + ch - 48;
    ch = getchar();
  }
  return s * f;
}
constexpr int N = 1e5 + 100;
int dp[2500][2500];
int nums[2500][26], numt[2500][26];
void solve() {
  int n = read<int>();
  string s, t;
  cin >> s >> t;
  s = "*" + s, t = "*" + t;
  for (int i = 1; i <= n; ++i) {
    int ids = s[i] - 'a', idt = t[i] - 'a';
    for (int j = 0; j < 26; ++j) {
      nums[i][j] = nums[i - 1][j] + (j == ids);
      numt[i][j] = numt[i - 1][j] + (j == idt);
    }
  }
  for (int i = 0; i < 26; ++i) {
    if (nums[n][i] != numt[n][i]) {
      cout << -1 << '\n';
      return;
    }
  }
  for (int i = 0; i <= n; i++) {
    dp[i][0] = dp[0][i] = 0;
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      dp[i][j] = -0x3f3f3f3f;
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = i; j <= n; ++j) {
      if (s[i] == t[j]) {
        bool flag = false;
        for (int z = 0; z < 26; ++z) {
          if (nums[i][z] > numt[j][z]) flag = true;
        }
        if (not flag) dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + 1);
      }
      if (j - 1 >= i) dp[i][j] = max(dp[i][j], dp[i][j - 1]);
      dp[i][j] = max(dp[i - 1][j], dp[i][j]);
    }
  }
  cout << n - dp[n][n] << '\n';
}
int main() {
  int _ = read<int>();
  while (_--) solve();
}
