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;
using INT = long long;
char A[2222], B[2222];
int cnt[33];
int SA[2222][26], SB[2222][30];
int dp[2222][2222];
int solve(int n) {
  for (int i = 0; i <= 26; i++)
    if (cnt[i]) return -1;
  for (int i = 0; i <= n + 2; i++)
    for (int j = 0; j <= n + 2; j++) dp[i][j] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= 26; j++)
      SA[i][j] = SA[i - 1][j], SB[i][j] = SB[i - 1][j];
    SA[i][A[i] - 'a']++;
    SB[i][B[i] - 'a']++;
  }
  for (int i = 0; i <= n; i++) dp[0][i] = 0, dp[i][0] = -0x3f3f3f3f;
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      if (SA[i][A[i] - 'a'] > SB[j][A[i] - 'a']) {
        dp[i][j] = -0x3f3f3f3f;
        continue;
      }
      if (i > j) {
        dp[i][j] = -0x3f3f3f3f;
        continue;
      }
      dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
      if (A[i] == B[j]) dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + 1);
    }
  return n - dp[n][n];
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    scanf("%s %s", A + 1, B + 1);
    memset(cnt, 0, sizeof cnt);
    for (int i = 1; i <= n; i++) {
      cnt[A[i] - 'a']++;
      cnt[B[i] - 'a']--;
    }
    printf("%d\n", solve(n));
  }
  return 0;
}
