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<bits/stdc++.h>
using namespace std;
int n,s;
int fa[200200];
int d[200200];
int rt1,rt2;

int te;
struct data{
	int to,nxt,id;
}mp[1000100];
int head[200200],cnt;
void link(int x,int y,int id)
{
	mp[++cnt].to=y;
	mp[cnt].id=id;
	mp[cnt].nxt=head[x];
	head[x]=cnt;
}

int ans[200200];
bool del[500500];
int sta[500500][2],top;
int ro[500500],cr;
void dfs(int s)
{
	sta[++top][0]=s;
	sta[top][1]=1;
	int x,v;
	while(top)
	{
		x=sta[top][0];
		int fl=1;
		if(sta[top][1]==1)
		{
			for(int i=head[x];i;i=mp[i].nxt)
			{
				if(del[mp[i].id])continue;
				del[mp[i].id]=1;
				head[x]=mp[i].nxt;
				fl=0;
				v=mp[i].to;
				sta[++top][0]=v;
				sta[top][1]=1;
				break;
			}
			if(fl)
			{
				sta[top][1]=2;
				continue;
			}
		}
		else if(sta[top][1]==2)
		{
			ro[++cr]=sta[top][0];
			top--;
		}
	}
}
int main()
{
	scanf("%d",&n);
	s=n*2+1;
	for(int i=1;i<=n;++i)
	{
		scanf("%d",&fa[i]);
		if(fa[i]==-1)fa[i]=s;
		d[fa[i]]++;
		++te;
		link(fa[i],i,te);
		link(i,fa[i],te);
	}
	for(int i=1;i<=n;++i)
	{
		scanf("%d",&fa[i]);
		if(fa[i]==-1)fa[i]=s;
		else fa[i]=s-fa[i];
		d[fa[i]]++;
		++te;
		link(fa[i],s-i,te);
		link(s-i,fa[i],te);
	}
	for(int i=1;i<s;++i)d[i]=(!(d[i]&1));
	for(int i=1;i<=n;++i)
	{
		if(d[i]!=d[s-i])
		{
			puts("IMPOSSIBLE");
			return 0;
		}
		if(d[i])
		{
			++te;
			link(i,s-i,te);
			link(s-i,i,te);
		}
	}
	dfs(1);
	puts("POSSIBLE");
	for(int i=cr;i>1;--i)
		if(ro[i]+ro[i-1]==s)
		{
			if(ro[i]<=n)ans[ro[i]]=ans[ro[i-1]]=1;
			else ans[ro[i]]=ans[ro[i-1]]=-1;
		}
	for(int i=1;i<=n;++i)printf("%d ",ans[i]);puts("");
	return 0;
}
