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 P = 1e9 + 7, INF = 0x3f3f3f3f;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long qpow(long long a, long long n) {
  long long r = 1 % P;
  for (a %= P; n; a = a * a % P, n >>= 1)
    if (n & 1) r = r * a % P;
  return r;
}
long long inv(long long first) {
  return first <= 1 ? 1 : inv(P % first) * (P - P / first) % P;
}
inline int rd() {
  int first = 0;
  char p = getchar();
  while (p < '0' || p > '9') p = getchar();
  while (p >= '0' && p <= '9') first = first * 10 + p - '0', p = getchar();
  return first;
}
const int N = 2e3 + 10;
int n, dp[N][N], suf[N][26], suf2[N][26];
char s[N], t[N];
void chkmin(int &a, int b) { a > b ? a = b : 0; }
void work() {
  scanf("%d%s%s", &n, s + 1, t + 1);
  for (int i = 1; i <= n; ++i) s[i] -= 'a', t[i] -= 'a';
  memset(suf[n + 1], 0, sizeof suf[0]);
  memset(suf2[n + 1], 0, sizeof suf2[0]);
  for (int i = n; i >= 1; --i) {
    memcpy(suf[i], suf[i + 1], sizeof suf[0]);
    memcpy(suf2[i], suf2[i + 1], sizeof suf2[0]);
    ++suf[i][s[i]], ++suf2[i][t[i]];
  }
  for (int i = 0; i <= n; ++i)
    for (int j = i; j <= n; ++j) dp[i][j] = INF;
  dp[0][0] = 0;
  for (int i = 0; i <= n; ++i)
    for (int j = i; j <= n; ++j) {
      int &r = dp[i][j];
      if (r == INF) continue;
      if (s[i + 1] == t[j + 1]) chkmin(dp[i + 1][j + 1], r);
      if (j < n && suf[i + 1][t[j + 1]] > suf2[j + 2][t[j + 1]])
        chkmin(dp[i][j + 1], r);
      chkmin(dp[i + 1][j], r + 1);
    }
  if (dp[n][n] == INF) dp[n][n] = -1;
  printf("%d\n", dp[n][n]);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) work();
}
