CodeForces - 240D. Merging Two Decks 贪心+暴力

There are two decks of cards lying on the table in front of you, some cards in these decks lay face up, some of them lay face down. You want to merge them into one deck in which each card is face down. You're going to do it in two stages.

The first stage is to merge the two decks in such a way that the relative order of the cards from the same deck doesn't change. That is, for any two different cards iand j in one deck, if card i lies above card j, then after the merge card i must also be above card j.

The second stage is performed on the deck that resulted from the first stage. At this stage, the executed operation is the turning operation. In one turn you can take a few of the top cards, turn all of them, and put them back. Thus, each of the taken cards gets turned and the order of these cards is reversed. That is, the card that was on the bottom before the turn, will be on top after it.

Your task is to make sure that all the cards are lying face down. Find such an order of merging cards in the first stage and the sequence of turning operations in the second stage, that make all the cards lie face down, and the number of turns is minimum.

Input

The first input line contains a single integer n — the number of cards in the first deck (1 ≤ n ≤ 105).

The second input line contains n integers, separated by single spaces a1, a2, ..., an(0 ≤ ai ≤ 1). Value ai equals 0, if the i-th card is lying face down, and 1, if the card is lying face up. The cards are given in the order from the topmost one to the bottommost one.

The third input line contains integer m — the number of cards in the second deck (1 ≤ m ≤ 105).

The fourth input line contains m integers, separated by single spaces b1, b2, ..., bm(0 ≤ bi ≤ 1). Value bi equals 0, if the i-th card is lying face down, and 1, if the card is lying face up. The cards are given in the order from the topmost to the bottommost.

Output

In the first line print n + m space-separated integers — the numbers of the cards in the order, in which they will lie after the first stage. List the cards from top to bottom. The cards from the first deck should match their indexes from 1 to n in the order from top to bottom. The cards from the second deck should match their indexes, increased by n, that is, numbers from n + 1 to n + m in the order from top to bottom.

In the second line print a single integer x — the minimum number of turn operations you need to make all cards in the deck lie face down. In the third line print xintegers: c1, c2, ..., cx (1 ≤ ci ≤ n + m), each of them represents the number of cards to take from the top of the deck to perform a turn operation. Print the operations in the order, in which they should be performed.

If there are multiple optimal solutions, print any of them. It is guaranteed that the minimum number of operations doesn't exceed 6·105.

Examples

Input

3
1 0 1
4
1 1 1 1

Output

1 4 5 6 7 2 3
3
5 6 7

Input

5
1 1 1 1 1
5
0 1 0 1 0

Output

6 1 2 3 4 5 7 8 9 10
4
1 7 8 9

题意：

有两叠扑克牌，我们首先要将他们合并在一起，每张扑克牌都有0 1两面。通过反转可以将0 与 1交换，问最少多少操作可以将扑克牌的0全部朝上，合并规则：保持每一叠的相对顺序不变，反转的规则是：从当前位置开始前面的全部都反转？

分析：

我们发现一个策略就是，因为只能翻转前面的，我们需要保证前面的相同，01010，我们遇到不同的就翻转则保证最小次数。

01010

11010

00010

11110

00000

对于最后一个为1的话，默认a[n+1]=0即可

我们保证他们的相同的状态尽量在一起即可。所以我们就以0为第一个或者以1为第一个枚举即可

#include<bits/stdc++.h>
const int N = 500000+5;
using namespace std;
int n,m;
int a[N];
int vis0[N],vis1[N];
vector<int> ans,ans0,ans1;
int res[N];
void judge(int sta,int vis[])
{
    memset(vis,0,sizeof(vis));
    int cnt=1;
    int i=1,j=n+1;
    while(i<=n||j<=n+m)
    {

        while(i<=n&&a[i]==sta)
        {
            vis[cnt++]=i++;
        }
        while(j<=n+m&&a[j]==sta)
        {
            vis[cnt++]=j++;
        }
        sta=sta^1;

    }
}

int main()
{
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    scanf("%d",&n);
    for(int i=1; i<=n; i++)
        scanf("%d",&a[i]);

    scanf("%d",&m);
    for(int i=n+1; i<=n+m; i++)
        scanf("%d",&a[i]);


    judge(0,vis0);
    for(int i=1; i<=n+m; i++)
    {
        if(a[vis0[i]]!=a[vis0[i+1]])
            ans0.push_back(i);
    }

    judge(1,vis1);
    for(int i=1; i<=n+m; i++)
    {
        if(a[vis1[i]]!=a[vis1[i+1]])
            ans1.push_back(i);
    }

    if(ans1.size()<=ans0.size())
    {
        for(int i=1; i<=n+m; i++)
            printf("%d ",vis1[i]);

        ans=ans1;
    }
    else
    {
        for(int i=1; i<=n+m; i++)
            printf("%d ",vis0[i]);
        ans=ans0;
    }



    printf("\n");
    printf("%d\n",ans.size());
    for(int i=0; i<ans.size(); i++)
        printf("%d ",ans[i]);

    printf("\n");

    return 0;
}