#include <iostream>
#include <iomanip>
#include <cstring>
#include <functional>
using namespace std;

const int N = 105;
double p[N], q[N];
double w[N][N], dp[N][N];
int root[N][N];
int n;

// 输出先序遍历
void preorder(int l, int r) {
    if (l > r) return;
    int k = root[l][r];
    cout << k;
    // 判断后面是否还有输出
    if (!(l == 1 && r == n))
        cout << " ";
    preorder(l, k - 1);
    preorder(k + 1, r);
}

int main() {

    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> p[i];
    for (int i = 0; i <= n; i++)
        cin >> q[i];
    // 初始化空区间
    for (int i = 1; i <= n + 1; i++) {
        dp[i][i - 1] = q[i - 1];
        w[i][i - 1] = q[i - 1];
    }

    // 区间DP
    for (int len = 1; len <= n; len++) {
        for (int l = 1; l + len - 1 <= n; l++) {
            int r = l + len - 1;
            // 计算 w[l][r]
            w[l][r] = w[l][r - 1] + p[r] + q[r];
            dp[l][r] = 1e18;
            // 枚举根
            for (int k = l; k <= r; k++) {
                double cost =
                    dp[l][k - 1]
                    + dp[k + 1][r]
                    + w[l][r];
                // 字典序最小：相等时取更小根
                if (cost < dp[l][r]) {
                    dp[l][r] = cost;
                    root[l][r] = k;
                }
            }
        }
    }

    // 输出先序遍历
    bool first = true;

    function<void(int,int)> dfs = [&](int l, int r) {
        if (l > r) return;
        int k = root[l][r];
        if (!first) cout << " ";
        first = false;
        cout << k;
        dfs(l, k - 1);
        dfs(k + 1, r);
    };

    dfs(1, n);
    return 0;
}