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 INF = 1e9;
const int maxn = 2e3 + 10;
char s[maxn], t[maxn];
int N;
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &N);
    scanf("%s", s + 1);
    scanf("%s", t + 1);
    vector<vector<int>> cnts(N + 1, vector<int>(26));
    vector<vector<int>> cntt(N + 1, vector<int>(26));
    for (int i = 1; i <= N; ++i) {
      for (int j = 0; j < 26; ++j) {
        cnts[i][j] = cnts[i - 1][j] + ((s[i] - 'a') == j);
        cntt[i][j] = cntt[i - 1][j] + ((t[i] - 'a') == j);
      }
    }
    int ok = 1;
    for (int j = 0; j < 26; ++j) {
      if (cnts[N][j] != cntt[N][j]) {
        ok = 0;
      }
    }
    if (!ok) {
      printf("-1\n");
      continue;
    }
    vector<vector<int>> DP(N + 1, vector<int>(N + 1));
    for (int i = 1; i <= N; ++i) {
      for (int j = 1; j <= N; ++j) {
        DP[i][j] = max(DP[i - 1][j], DP[i][j - 1]);
        if (s[i] == t[j]) {
          int ok2 = 1;
          for (int k = 0; k < 26; ++k) {
            if (cnts[N][k] - cnts[i - 1][k] < cntt[N][k] - cntt[j - 1][k]) {
              ok2 = 0;
            }
          }
          if (ok2) {
            DP[i][j] = max(DP[i][j], DP[i - 1][j - 1] + 1);
          }
        }
      }
    }
    printf("%d\n", N - DP[N][N]);
  }
}
