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 ckmn(T &a, T b) {
  a = min(a, b);
}
template <typename T>
void ckmx(T &a, T b) {
  a = max(a, b);
}
int ri() {
  int x;
  scanf("%i", &x);
  return x;
}
void rd() {}
template <typename... T>
void rd(int &x, T &...args) {
  scanf("%i", &x);
  rd(args...);
}
template <typename... T>
void rd(long long &x, T &...args) {
  scanf("%lld", &x);
  rd(args...);
}
template <typename... T>
void rd(double &x, T &...args) {
  scanf("%lf", &x);
  rd(args...);
}
template <typename... T>
void rd(pair<int, int> &x, T &...args) {
  scanf("%i %i", &x.first, &x.second);
  rd(args...);
}
const int N = 2050;
const int inf = 1e9 + 7;
char s[N], t[N];
int c_s[26][N], c_t[26][N];
int main() {
  for (int tc = ri(); tc--;) {
    int n = ri();
    scanf("%s", s + 1);
    scanf("%s", t + 1);
    for (int i = 0; i < 26; i++) c_s[i][n + 1] = c_t[i][n + 1] = 0;
    for (int i = n; i >= 1; i--) {
      for (int j = 0; j < 26; j++)
        c_s[j][i] = c_s[j][i + 1], c_t[j][i] = c_t[j][i + 1];
      c_s[s[i] - 'a'][i]++;
      c_t[t[i] - 'a'][i]++;
    }
    vector<vector<int>> dp(n + 1, vector<int>(n + 1, inf));
    dp[n][n] = 0;
    for (int i = n; i >= 0; i--) {
      for (int j = n; j >= 0; j--) {
        if (i && j && s[i] == t[j]) ckmn(dp[i - 1][j - 1], dp[i][j]);
        if (i) ckmn(dp[i - 1][j], dp[i][j] + 1);
        if (j) {
          int now = t[j] - 'a';
          if (c_s[now][i + 1] > c_t[now][j + 1]) ckmn(dp[i][j - 1], dp[i][j]);
        }
      }
    }
    if (dp[0][0] == inf)
      printf("-1\n");
    else
      printf("%i\n", dp[0][0]);
  }
  return 0;
}
