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<bits/stdc++.h>
using namespace std;

const int MAXN = 20, MAXS = (1 << 18) + 20, INF = 0x3f3f3f3f;
int n, a[MAXN], b[MAXN], dp[MAXS][MAXN];

void umin(int &x, int y) { if(y < x) x = y; }
int main() {
    memset(dp, 0x3f, sizeof dp);
    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]);
    a[n] = b[n] = 0;
    dp[0][n] = 0;
    int full = (1 << n) - 1;
    for(int S = 0; S < full; ++S) {
        int pp = __builtin_popcount(S), ptr1 = 0, ptr2 = pp;
        static int idx[MAXN];
        for(int i = 0; i < n; ++i) {
            if(S >> i & 1) idx[i] = ptr1++;
            else idx[i] = ptr2++;
        }
        for(int i = 0; i <= n; ++i) if(dp[S][i] != INF) {
            int las = ((pp & 1) ^ (i & 1)) ? a[i] : b[i];
            for(int j = 0; j < n; ++j) if(~S >> j & 1) {
                int cur = ((pp & 1) ^ (j & 1)) ? b[j] : a[j];
                if(cur < las) continue;
                umin(dp[S ^ (1 << j)][j], dp[S][i] + idx[j] - pp);
            }
        }
    }
    int ans = *min_element(dp[full], dp[full] + n);
    printf("%d\n", ans == INF ? -1 : ans);
    return 0;
}