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 A = 26;
const int N = 2010;
const int INF = 696969696;
char a[N], b[N];
int t, n, f[N][N];
vector<int> cntA[N], cntB[N];
int main() {
  for (int i = 0; i < N; ++i) {
    cntA[i].resize(A, 0), cntB[i].resize(A, 0);
  }
  cin >> t;
  while (t--) {
    scanf("%d %s %s", &n, a + 1, b + 1);
    cntA[n + 1].resize(A, 0), cntB[n + 1].resize(A, 0);
    for (int i = n; i >= 1; --i) {
      cntA[i] = cntA[i + 1], cntB[i] = cntB[i + 1];
      ++cntA[i][a[i] - 'a'], ++cntB[i][b[i] - 'a'];
    }
    if (cntA[1] != cntB[1]) {
      puts("-1");
      continue;
    }
    for (int i = 0; i <= n; ++i)
      for (int j = 0; j <= n; ++j) {
        if (i + j == 0) {
          f[i][j] = 0;
          continue;
        }
        f[i][j] = INF;
        if (!j) continue;
        if (i) {
          f[i][j] = min(f[i][j], 1 + f[i - 1][j]);
          if (a[i] == b[j]) f[i][j] = min(f[i][j], f[i - 1][j - 1]);
        }
        if (cntA[i + 1][b[j] - 'a'] > cntB[j + 1][b[j] - 'a']) {
          f[i][j] = min(f[i][j], f[i][j - 1]);
        }
      }
    int ans = f[n][n] > INF / 2 ? -1 : f[n][n];
    printf("%d\n", ans);
  }
  return 0;
}
