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<cstring>
#include<algorithm>
typedef long long ll;
const int N=1e6+5;
const int mod=1e9+7;
struct node{
    int u,v,nxt;
};
struct Graph{
    node edge[N*2];
    int head[N];
    int mcnt;
    Graph(){memset(head,0,sizeof head);mcnt=0;}
    void add_edge(int u,int v){
        mcnt++;
        edge[mcnt].u=u;
        edge[mcnt].v=v;
        edge[mcnt].nxt=head[u];
        head[u]=mcnt;
    }
}G1,G2,G;
int n;
int f1[N],f2[N],a1[N],a2[N];
int rt1,rt2;
int s[N],top;
void dfs(Graph *g,int u){
    for(int i=g->head[u];i;i=g->edge[i].nxt){
        int v=g->edge[i].v;
        dfs(g,v);
    }
    if(a1[u])
        s[++top]=u;
    while(top>=2){
        G.add_edge(s[top],s[top-1]);
        G.add_edge(s[top-1],s[top]);
        top-=2;
    }
}
int color[N];
void dfs(int u,int c){
    color[u]=c;
    for(int i=G.head[u];i;i=G.edge[i].nxt){
        int v=G.edge[i].v;
        if(color[v]==0)
            dfs(v,-c);
    }
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        a1[i]=a2[i]=1;
    for(int i=1;i<=n;i++){
        scanf("%d",&f1[i]);
        if(f1[i]==-1)
            rt1=i;
        else{
            G1.add_edge(f1[i],i);
            a1[f1[i]]^=1;
        }
    }
    for(int i=1;i<=n;i++){
        scanf("%d",&f2[i]);
        if(f2[i]==-1)
            rt2=i;
        else{
            G2.add_edge(f2[i],i);
            a2[f2[i]]^=1;
        }
    }
    for(int i=1;i<=n;i++)
        if(a1[i]!=a2[i]){
            puts("IMPOSSIBLE");
            return 0;
        }
    puts("POSSIBLE");
    top=0;
    dfs(&G1,rt1);
    top=0;
    dfs(&G2,rt2);
    for(int i=1;i<=n;i++)
        if(color[i]==0)
            dfs(i,1);
    for(int i=1;i<=n;i++)
        if(!a1[i])
            printf("0 ");
        else
            printf("%d ",color[i]);
}
