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;
long long gcd(long long a, long long b) {
  for (; b; a %= b, swap(a, b))
    ;
  return a;
}
const int INF = 987654321;
int n;
string s, t;
int ps1[26][2001];
int ps2[26][2001];
int dp[2001][2001];
int solve(int i, int j) {
  if (i == 0) return 0;
  int& ret = dp[i][j];
  if (ret != -1) return ret;
  ret = solve(i - 1, j) + 1;
  if (s[i - 1] == t[j - 1]) ret = min(ret, solve(i - 1, j - 1));
  int idx = t[j - 1] - 'a';
  int s1 = ps1[idx][n] - ps1[idx][i];
  int s2 = ps2[idx][n] - ps2[idx][j];
  if (s1 > s2) ret = min(ret, solve(i, j - 1));
  return ret;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int T;
  cin >> T;
  while (T--) {
    cin >> n;
    cin >> s >> t;
    int cnt[2][26] = {
        0,
    };
    for (int i = 0; i < n; i++) {
      cnt[0][s[i] - 'a']++;
      cnt[1][t[i] - 'a']++;
    }
    bool flag = false;
    for (int i = 0; i < 26; i++) {
      if (cnt[0][i] != cnt[1][i]) {
        flag = true;
        break;
      }
    }
    if (flag) {
      cout << "-1\n";
      continue;
    }
    for (int i = 0; i <= n; i++)
      for (int j = 0; j <= n; j++) dp[i][j] = -1;
    int s1[26] = {
        0,
    };
    int s2[26] = {
        0,
    };
    for (int i = 0; i < n; i++) {
      s1[s[i] - 'a']++;
      s2[t[i] - 'a']++;
      for (int j = 0; j < 26; j++) {
        ps1[j][i + 1] = s1[j];
        ps2[j][i + 1] = s2[j];
      }
    }
    cout << solve(n, n) << '\n';
  }
}
