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<functional>
#include<algorithm>
#include<iterator>
#include<cstring>
#include<cstdio>
using namespace std;
void GG()
{
	puts("IMPOSSIBLE");
	exit(0);
}
int da[111111],db[111111],pa[111111],pb[111111];
int n;
struct R{int v,z;}e[1111111];
int K[222222],cur[222222],ce=1;
void ins(int a,int b){e[++ce]=(R){b,K[a]};K[a]=ce;}
int viz[1111111],res[111111];
const int s=0;
int d;
void dfs(int u)
{
	++d;
	for(int&i=cur[u];i;i=e[i].z)if(!viz[i])
	{
		viz[i]=1;
		viz[i^1]=-1;
		dfs(e[i].v);
	}
}
void euler()
{
	memcpy(cur,K,sizeof(cur));
	dfs(s);
	puts("POSSIBLE");
	for(int u=1;u<=n;++u)
	{
		int a=0;
		for(int i=K[u];i;i=e[i].z)
		{
			if(e[i].v==pa[u])
				res[u]+=viz[i],
				res[pa[u]]-=viz[i];
		}
	}
	for(int i=1;i<=n;++i)printf("%d ",res[i]);
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;++i)
	{
		scanf("%d",&pa[i]);
		if(pa[i]==-1)pa[i]=s;
		ins(pa[i],i),ins(i,pa[i]);
		++da[pa[i]];
	}
	for(int i=1;i<=n;++i)
	{
		scanf("%d",&pb[i]);
		if(pb[i]==-1)pb[i]=s;
		ins(pb[i]+n+1,i+n+1),ins(i+n+1,pb[i]+n+1);
		++db[pb[i]];
	}
	ins(s,s+n+1);
	ins(s+n+1,s);
	for(int i=1;i<=n;++i)
	{
		if((da[i]&1)!=(db[i]&1))
			GG();
		#define gmin(a,b) ((a)<(b)?(a):(b))
		int z=gmin(da[i],db[i])+1;
		for(int j=1;j<=z;++j)ins(i,i+n+1),ins(i+n+1,i);
	}
	euler();
	return 0;
}