Problem: We have N cards numbered 1, 2, ..., N. Card i (1 \leq i \leq N) has an integer A_i written in red ink on one side and an integer B_i written in blue ink on the other side. Initially, these cards are arranged from left to right in the order from Card 1 to Card N, with the red numbers facing up.

Determine whether it is possible to have a non-decreasing sequence facing up from left to right (that is, for each i (1 \leq i \leq N - 1), the integer facing up on the (i+1)-th card from the left is not less than the integer facing up on the i-th card from the left) by repeating the operation below. If the answer is yes, find the minimum number of operations required to achieve it.

* Choose an integer i (1 \leq i \leq N - 1). Swap the i-th and (i+1)-th cards from the left, then flip these two cards.

Constraints

* 1 \leq N \leq 18
* 1 \leq A_i, B_i \leq 50 (1 \leq i \leq N)
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
A_1 A_2 ... A_N
B_1 B_2 ... B_N


Output

If it is impossible to have a non-decreasing sequence, print `-1`. If it is possible, print the minimum number of operations required to achieve it.

Examples

Input

3
3 4 3
3 2 3


Output

1


Input

2
2 1
1 2


Output

-1


Input

4
1 2 3 4
5 6 7 8


Output

0


Input

5
28 15 22 43 31
20 22 43 33 32


Output

-1


Input

5
4 46 6 38 43
33 15 18 27 37


Output

3

Solution (in CPP):

Solution:
#include <cstdio>
#include <utility>
#include <algorithm>

int n;
int A[18];
int B[18];
int C[18];
int D[18];

/*
dp[S][k] = OR dp[S-i][A[i]]
*/

int dp[1<<18][51];

int main(){
  scanf("%d", &n);
  for(int i = 0; i < n; i++) scanf("%d", A+i);
  for(int i = 0; i < n; i++) scanf("%d", B+i);

  for(int i = 1; i < (1 << n); i++){
    for(int k = 1; k <= 50; k++) dp[i][k] = 1000000000;
    for(int j = i; j; j &= j-1){
      int v = __builtin_ctz(j);
      bool t = __builtin_parity(i & ~((1<<(v+1))-1));
      int l = __builtin_popcount(~i & ((1<<v)-1));
      if(t^(l&1)) for(int k=1;k<=B[v];k++) dp[i][B[v]] = std::min(dp[i][B[v]], dp[i ^ (1 << v)][k] + l);
      else  for(int k=1;k<=A[v];k++) dp[i][A[v]] = std::min(dp[i][A[v]], dp[i ^ (1 << v)][k] + l);
    }
//for(int k = 1; k <= 50; k++) printf("%d ", dp[i][k]);
//puts("");

  }
  int ans = 1000000000;
  for(int k = 1; k <= 50; k++) ans = std::min(ans, dp[(1<<n)-1][k]);

  if(ans == 1000000000) puts("-1");
  else printf("%d\n", ans);


  return 0;
}