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;
template <typename T>
inline T abs(T t) {
  return t < 0 ? -t : t;
}
const long long modn = 1000000007;
inline long long mod(long long x) { return x % modn; }
const int MAXN = 2123;
const int INF = 0x3f3f3f3f;
int n, m, k;
char s[MAXN];
char ss[MAXN];
int sum[30][MAXN], summ[30][MAXN];
int dp[MAXN][MAXN];
int go(int i, int j) {
  if (i < 0 || j < 0) return 0;
  int &r = dp[i][j];
  if (r != -1) return r;
  r = INF;
  if (s[i] == ss[j]) r = min(r, go(i - 1, j - 1));
  if (sum[ss[j] - 'a'][i + 1] > summ[ss[j] - 'a'][j + 1])
    r = min(r, go(i, j - 1));
  r = min(r, go(i - 1, j) + 1);
  return r;
}
int main() {
  int t;
  scanf("%d", &t);
  for (int tt = 1; tt <= t; tt++) {
    scanf("%d", &n);
    for (int a = 0; a <= n; a++)
      for (int b = 0; b <= n; b++) dp[a][b] = -1;
    scanf(" %s", s);
    scanf(" %s", ss);
    for (int c = 0; c < 26; c++) {
      for (int a = n - 1; a >= 0; a--) {
        sum[c][a] = sum[c][a + 1] + (s[a] == 'a' + c);
        summ[c][a] = summ[c][a + 1] + (ss[a] == 'a' + c);
      }
    }
    bool ok = true;
    for (int c = 0; c < 26; c++) {
      if (sum[c][0] != summ[c][0]) {
        ok = false;
      }
    }
    if (!ok) {
      puts("-1");
      continue;
    }
    printf("%d\n", go(n - 1, n - 1));
  }
}
