There are two rooted trees, each with N vertices. The vertices of each tree are numbered 1 through N. In the first tree, the parent of Vertex i is Vertex A_i. Here, A_i=-1 if Vertex i is the root of the first tree. In the second tree, the parent of Vertex i is Vertex B_i. Here, B_i=-1 if Vertex i is the root of the second tree.

Snuke would like to construct an integer sequence of length N, X_1 , X_2 , ... , X_N, that satisfies the following condition:

* For each vertex on each tree, let the indices of its descendants including itself be a_1 , a_2 , ..., a_k. Then, abs(X_{a_1} + X_{a_2} + ... + X_{a_k})=1 holds.



Determine whether it is possible to construct such a sequence. If the answer is possible, find one such sequence.

Constraints

* 1 \leq N \leq 10^5
* 1 \leq A_i \leq N, if Vertex i is not the root in the first tree.
* A_i = -1, if Vertex i is the root in the first tree.
* 1 \leq B_i \leq N, if Vertex i is not the root in the second tree.
* B_i = -1, if Vertex i is the root in the second tree.
* Input corresponds to valid rooted trees.

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 not possible to construct an integer sequence that satisfies the condition, print `IMPOSSIBLE`. If it is possible, print `POSSIBLE` in the first line. Then, in the second line, print X_1 , X_2 , ... , X_N, an integer sequence that satisfies the condition.

Examples

Input

5
3 3 4 -1 4
4 4 1 -1 1


Output

POSSIBLE
1 -1 -1 3 -1


Input

6
-1 5 1 5 1 3
6 5 5 3 -1 3


Output

IMPOSSIBLE


Input

8
2 7 1 2 2 1 -1 4
4 -1 4 7 4 4 2 4


Output

POSSIBLE
1 2 -1 0 -1 1 0 -1
#include <cstdio>
#include <cctype>
#include <cstring>
#define repu(i,x,y) for (int i=x; i<=y; ++i)
using namespace std;

int n,root,e[100100][2],ans[100100];
struct edge
{
    int v;
    edge *nxt;
} pool[100100],*tp,*fst[100100];

int getint()
{
    char ch;
    while (!isdigit(ch=getchar()) && ch!='-');
    bool flag=ch=='-';
    if (flag)
        ch=getchar();
    int x=ch-'0';
    for (; isdigit(ch=getchar()); x=x*10+ch-'0');
    return flag?-x:x;
}

void addedge(int u,int v)
{
    (e[u][0]?e[u][1]:e[u][0])=v;
    (e[v][0]?e[v][1]:e[v][0])=u;
}

int dfs1(int x)
{
    int ret=0;
    for (edge *i=fst[x]; i; i=i->nxt)
    {
        int t=dfs1(i->v);
        if (ret)
            addedge(ret,t),ret=0;
        else
            ret=t;
    }
    return ret?ret:x;
}

void dfs2(int x,int w)
{
    ans[x]=w;
    repu(i,0,1)
        if (!ans[e[x][i]])
            dfs2(e[x][i],-w);
}

int main()
{
    n=getint();
    repu(j,1,2)
    {
        memset(fst,0,sizeof(fst)),tp=pool;
        repu(i,1,n)
        {
            int fa=getint();
            if (fa==-1)
                root=i;
            else
                *tp=(edge){i,fst[fa]},fst[fa]=tp++;
        }
        int t=dfs1(root);
        addedge(n+1,t);
    }
    repu(i,1,n)
        if (e[i][0])
        {
            if (!e[i][1])
            {
                puts("IMPOSSIBLE");
                return 0;
            }
            if (!ans[i])
                dfs2(i,1);
        }
    puts("POSSIBLE");
    repu(i,1,n)
        printf("%d ",ans[i]);
    return 0;
}