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 maxn = 2e3 + 7;
const int INF = 1e9 + 7;
int n;
string s, t;
int dp[maxn][maxn];
int vis1[30], vis2[30];
int main() {
  int ti;
  scanf("%d", &ti);
  while (ti--) {
    cin >> n >> s >> t;
    swap(s, t);
    reverse(s.begin(), s.end());
    reverse(t.begin(), t.end());
    for (int i = 0; i <= n; i++)
      for (int j = 0; j <= n; j++) dp[i][j] = INF;
    memset(vis1, 0, sizeof(vis1));
    for (int i = 0; i <= n; i++) {
      if (i >= 1) vis1[s[i - 1] - 'a']++;
      for (int j = 0; j <= n; j++) {
        if (j != 0)
          vis2[t[j - 1] - 'a']++;
        else
          memset(vis2, 0, sizeof(vis2));
        if (j < i) continue;
        if (i == 0)
          dp[i][j] = j - i;
        else {
          if (s[i - 1] == t[j - 1]) dp[i][j] = min(dp[i][j], dp[i - 1][j - 1]);
          dp[i][j] = min(dp[i][j], dp[i][j - 1] + 1);
          if (vis2[s[i - 1] - 'a'] >= vis1[s[i - 1] - 'a'])
            dp[i][j] = min(dp[i][j], dp[i - 1][j]);
        }
      }
    }
    if (dp[n][n] != INF)
      cout << dp[n][n] << endl;
    else
      cout << -1 << endl;
  }
  return 0;
}
