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>
void read(T &t) {
  t = 0;
  char ch = getchar();
  int f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  do {
    (t *= 10) += ch - '0';
    ch = getchar();
  } while ('0' <= ch && ch <= '9');
  t *= f;
}
const int INF = (1e9);
const int maxn = 2010;
int T, n;
char s[maxn], t[maxn];
int dp[maxn][maxn];
int c1[maxn][30], c2[maxn][30];
void update(int &x, int y) {
  if (x > y) x = y;
}
int main() {
  int flag, x;
  read(T);
  while (T--) {
    read(n);
    scanf("%s", s + 1);
    scanf("%s", t + 1);
    memset(c1[n + 1], 0, sizeof(c1));
    memset(c2[n + 1], 0, sizeof(c1));
    for (int i = n; i >= 1; i--) {
      memcpy(c1[i], c1[i + 1], sizeof(c1[i]));
      memcpy(c2[i], c2[i + 1], sizeof(c2[i]));
      c1[i][s[i] - 'a']++;
      c2[i][t[i] - 'a']++;
    }
    flag = 1;
    for (int i = 0; i < 26; i++)
      if (c1[1][i] != c2[1][i]) {
        flag = 0;
        break;
      }
    if (!flag) {
      printf("-1\n");
      continue;
    }
    for (int i = 0; i <= n; i++)
      for (int j = 0; j <= n; j++) dp[i][j] = INF;
    for (int j = 0; j <= n; j++) dp[0][j] = 0;
    for (int i = 1; i <= n; i++)
      for (int j = i; j <= n; j++) {
        x = t[j] - 'a';
        if (c1[i + 1][x] > c2[j + 1][x]) update(dp[i][j], dp[i][j - 1]);
        if (s[i] == t[j]) update(dp[i][j], dp[i - 1][j - 1]);
        update(dp[i][j], dp[i - 1][j] + 1);
      }
    printf("%d\n", dp[n][n]);
  }
  return 0;
}
