After solving Solution to the n Queens Puzzle by constructing, LoadingTime wants to solve a harder version of the N-Queen Problem. Some queens have been set on particular locations on the board in this problem. Can you help him??

Input

The input contains multiple test cases. Every line begins with an integer N (N<=50), then N integers followed, representing the column number of the queen in each rows. If the number is 0, it means no queen has been set on this row. You can assume there is at least one solution.

Output

For each test case, print a line consists of N numbers separated by spaces, representing the column number of the queen in each row. If there are more than one answer, print any one of them.

Example

Input:

4 0 0 0 0

8 2 0 0 0 4 0 0 0

Output:

2 4 1 3

2 6 1 7 4 8 3 5

#include<cstdio>

#include<cstring>

#include<string>

#include<algorithm>

using namespace std;

const int INF=1e9+;

const int ms=*;

const int maxn=ms*;

int N,ans[],res[];//ans存储第几个选择的编号，res保存第几行答案是第几列

struct DLX

{

    int n,id;

    int L[maxn],R[maxn],U[maxn],D[maxn];

    int C[maxn],S[maxn],loc[maxn][];

    int H[ms];

    void init(int nn=) //传列长

    {

        n=nn;

        for(int i=;i<=n;i++) U[i]=D[i]=i,L[i]=i-,R[i]=i+;

        L[]=n; R[n]=;

        id=n;

        memset(S,,sizeof(S));

        memset(H,-,sizeof(H));

    }

    void Link(int x,int y)

    {

        ++id;

        D[id]=y; U[id]=U[y];

        D[U[y]]=id; U[y]=id;

        loc[id][]=x,loc[id][]=y;

        C[id]=y; S[y]++;

        if(H[x]==-) H[x]=L[id]=R[id]=id;

        else

        {

            int a=H[x];

            int b=R[a];

            L[id]=a; R[a]=id;

            R[id]=b; L[b]=id;

            H[x]=id;

        }

    }

    void Remove(int c)

    {

        L[R[c]]=L[c];

        R[L[c]]=R[c];

        for(int i=D[c];i!=c;i=D[i])

            for(int j=R[i];j!=i;j=R[j])

        {

            U[D[j]]=U[j];

            D[U[j]]=D[j];

            S[C[j]]--;

        }

    }

    void Resume(int c)

    {

        for(int i=U[c];i!=c;i=U[i])

            for(int j=R[i];j!=i;j=R[j])

        {

            S[C[j]]++;

            U[D[j]]=j;

            D[U[j]]=j;

        }

        L[R[c]]=c;

        R[L[c]]=c;

    }

    bool dfs(int step)

    {

        if(step>=N) return true;

        if(R[]==) return false;

        int Min=INF,c=-;

        for(int i=R[];i;i=R[i])

        {

            if(i>N) break;

            if(Min>S[i]){ Min=S[i]; c=i; }

        }

        if(c==-) return false;

        Remove(c);

        for(int i=D[c];i!=c;i=D[i])

        {

            ans[step]=loc[i][];

            for(int j=R[i];j!=i;j=R[j]) Remove(C[j]);

            if(dfs(step+)) return true;

            for(int j=L[i];j!=i;j=L[j]) Resume(C[j]);

        }

        Resume(c);

        return false;

    }

}dlx;

bool vis[*];

int main()

{

    while(scanf("%d",&N)!=EOF)

    {

        dlx.init(N*-);

        memset(vis,false,sizeof(vis));

        int y;

        for(int x=;x<=N;x++)

        {

            scanf("%d",&y);

            if(y==) continue;

            int a=x,b=N+y,c=*N+N+x-y,d=*N+x+y-; //对应的行列斜对角编号

            vis[a]=vis[b]=vis[c]=vis[d]=true; //标记

            int t=(x-)*N+y-;

            dlx.Link(t,a); //连接

            dlx.Link(t,b);

            dlx.Link(t,c);

            dlx.Link(t,d);

        }

        for(int x=;x<=N;x++)

            for(int y=;y<=N;y++)

        {

            int a=x,b=N+y,c=*N+N+x-y,d=*N+x+y-;

            if(vis[a]||vis[b]||vis[c]||vis[d]) continue; //有被占据不考虑

            int t=(x-)*N+y-;

            dlx.Link(t,a);

            dlx.Link(t,b);

            dlx.Link(t,c);

            dlx.Link(t,d);

        }

        if(!dlx.dfs()) printf("No answer find\n");

        else

        {

            for(int i=;i<N;i++) res[ans[i]/N]=ans[i]%N;

            for(int i=;i<N;i++) printf("%d%c",res[i]+,i==N-?'\n':' ');

        }

    }

    return ;

}