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>
#include<cstring>
#include<algorithm>
#include<assert.h>
#define fo(i,a,b) for(int i=(a);i<=(b);++i)
#define fd(i,b,a) for(int i=(b);i>=(a);--i)
#define bfo(i,v,u) for(int i=BB[v],u=B[i][1];i;u=B[i=B[i][0]][1])
using namespace std;
typedef long long ll;
int read(){int n=0,p=1;char ch;for(ch=getchar();ch<'0' || ch>'9';ch=getchar())if(ch=='-') p=-1;for(;'0'<=ch && ch<='9';ch=getchar()) n=n*10+ch-'0';return n*p;}
const int N=2e5+5;
int n,S,B0,BB[N],B[N<<2][2],d0[N],d1[N],ans[N];
void link(int u,int v){B[++B0][1]=v,B[B0][0]=BB[u],BB[u]=B0;}
bool ban[N<<2];
void dfs(int v)
{
	for(int i=BB[v],u=B[i][1];i;i=BB[v],u=B[i][1])
	{
		BB[v]=B[i][0];
		if(ban[i]) continue;
		ban[i]=ban[i^1]=1;
		dfs(u);
		if(u==v+n) ans[v]=1;
		if(v==u+n) ans[u]=-1;
	}
}
int main()
{
	n=read();S=n+n+1;
	B0=1;
	fo(i,1,n)
	{
		int x=read();
		if(x==-1) link(S,i),link(i,S),++d0[i];
		else link(x,i),link(i,x),++d0[i],++d0[x];
	}
	fo(i,1,n)
	{
		int x=read();
		if(x==-1) link(S,n+i),link(n+i,S),++d1[i];
		else link(n+x,n+i),link(n+i,n+x),++d1[i],++d1[x];
	}
	fo(i,1,n)
		if((d0[i]+d1[i])&1) return puts("IMPOSSIBLE"),0;
		else
		if(d0[i]&1) link(i,n+i),link(n+i,i);
	puts("POSSIBLE");
	dfs(S);
	fo(i,1,n) printf("%d ",ans[i]);
	return 0;
}