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;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long q;
  cin >> q;
  while (q--) {
    long long n;
    cin >> n;
    string s, t;
    cin >> s >> t;
    vector<vector<long long>> scounter(n, vector<long long>(26, 0)),
        tcounter(n, vector<long long>(26, 0));
    scounter[n - 1][s[n - 1] - 'a']++;
    tcounter[n - 1][t[n - 1] - 'a']++;
    for (long long i = n - 2; i >= 0; i--) {
      for (long long j = 0; j < 26; j++) {
        scounter[i][j] = scounter[i + 1][j];
        tcounter[i][j] = tcounter[i + 1][j];
      }
      scounter[i][s[i] - 'a']++;
      tcounter[i][t[i] - 'a']++;
    }
    bool istrue = true;
    for (long long i = 0; i < 26; i++) {
      istrue = istrue & (scounter[0][i] == tcounter[0][i]);
    }
    tcounter.push_back(vector<long long>(26, 0));
    scounter.push_back(vector<long long>(26, 0));
    if (!istrue) {
      cout << -1 << '\n';
      continue;
    }
    vector<vector<long long>> dp(n, vector<long long>(n, 1e9));
    for (long long i = 0; i < n; i++) dp[0][i] = 0;
    for (long long i = 0; i < n; i++) {
      for (long long j = 0; j < n; j++) {
        if (min(i, j) - 1 >= 0 && s[i] == t[j])
          dp[i][j] = min(dp[i][j], dp[i - 1][j - 1]);
        if (j > 0 && scounter[i + 1][t[j] - 'a'] > tcounter[j + 1][t[j] - 'a'])
          dp[i][j] = min(dp[i][j], dp[i][j - 1]);
        if (i > 0) dp[i][j] = min(dp[i][j], dp[i - 1][j] + 1);
      }
    }
    cout << dp[n - 1][n - 1] << '\n';
  }
}
