Problem: 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

Solution (in CPP):

Solution:
#include<cstdio>
const int MaxN=100010;
int N,A[MaxN],B[MaxN],deg[MaxN*2],X[MaxN];
struct edge{int to;bool del;edge*rev,*next;}E[MaxN*12],*ne=E,*fir[MaxN*2];
void link(int u,int v,bool d=1){
	*ne=(edge){v,0,ne+1,fir[u]};fir[u]=ne++;
	*ne=(edge){u,0,ne-1,fir[v]};fir[v]=ne++;
	if(d)deg[u]++,deg[v]++;
}
void dfs(int i){
	while(fir[i]){
		edge*e=fir[i];
		fir[i]=e->next;
		if(e->del)continue;
		e->del=e->rev->del=1;
		if(i&&i<=N&&e->to>N)X[i]++;
		if(i>N&&e->to&&e->to<=N)X[e->to]--;
		dfs(e->to);
	}
}
int main(){
	scanf("%d",&N);
	for(int i=1;i<=N;i++){
		scanf("%d",A+i);
		link(A[i]<0?0:A[i],i);
	}
	for(int i=1;i<=N;i++){
		scanf("%d",B+i);
		link(B[i]<0?0:B[i]+N,i+N);
	}
	for(int i=1;i<=N;i++){
		if(deg[i]+deg[i+N]&1)return puts("IMPOSSIBLE"),0;
		for(int j=deg[i]>deg[i+N]?deg[i]:deg[i+N];j--;)link(i,i+N,0);
	}
	dfs(0);
	puts("POSSIBLE");
	for(int i=1;i<=N;i++)printf("%d%c",X[i]," \n"[i==N]);
}