Description

You are given two sequences of integer numbers. Write a program to determine their common increasing subsequence of maximal possible length.
Sequence S 1 , S 2 , . . . , S N of length N is called an increasing subsequence of a sequence A 1 , A 2 , . . . , A M of length M if there exist 1 <= i 1 < i 2 < . . . < i N <= M such that S j = A ij for all 1 <= j <= N , and S j < S j+1 for all 1 <= j < N .

Input

Each sequence is described with M --- its length (1 <= M <= 500) and M integer numbers A i (-2 31 <= A i < 2 31 ) --- the sequence itself.

Output

On the first line of the output file print L --- the length of the greatest common increasing subsequence of both sequences. On the second line print the subsequence itself. If there are several possible answers, output any of them.

Sample Input

5
1 4 2 5 -12
4
-12 1 2 4

Sample Output

2
1 4

Source

Northeastern Europe 2003, Northern Subregion

裸的LICS加上输出路径就把自己搞懵了；_； 一心想着记录父节点可以递归输出，然而远不用那么麻烦，最长递增公共子序列嘛==而且是二维的（废话），多加一个二维数组记录哈希了的前一个对应坐标，根据这个坐标记录每步的走法，这么说来，我原来递归的写法就只差一个数组呗，等回头补上

/*********
poj2127
2016.2.12
1168K	141MS	C++	1188B
*********/
#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int n,m,x[505],y[505],dp[505],pos[505][505],ans[505];
void solve()
{
    memset(dp,0,sizeof(dp));
    memset(pos,0,sizeof(pos));
    for(int i=1;i<=n;i++)
    {
        int tmp=0;
        memcpy(pos[i],pos[i-1],sizeof(pos[0]));///
        for(int j=1;j<=m;j++)
        {
            if(x[i]>y[j]&&dp[j]>dp[tmp]) tmp=j;
            if(x[i]==y[j]&&dp[tmp]+1>dp[j])
            {
                dp[j]=dp[tmp]+1;
                pos[i][j]=i*(m+1)+tmp;
            }
        }
    }
    int maxn=0;
    for(int i=1;i<=m;i++) if(dp[maxn]<dp[i]) maxn=i;
    int i=m*n+n+maxn;
    for(int j=dp[maxn];j>0;j--)
    {
        ans[j]=y[i%(m+1)];
        i=pos[i/(m+1)][i%(m+1)];
    }
    printf("%d\n",dp[maxn]);
    for(i=1;i<=dp[maxn];i++)
    {
        if(i!=1) printf(" %d",ans[i]);
        else printf("%d",ans[i]);
    } puts("");
}
int main()
{
   // freopen("cin.txt","r",stdin);
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++) scanf("%d",&x[i]);
        scanf("%d",&m);
        for(int i=1;i<=m;i++) scanf("%d",&y[i]);
        solve();
    }
    return 0;
}