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<set>
#include<map>
#include<deque>
#include<queue>
#include<stack>
#include<cmath>
#include<ctime>
#include<bitset>
#include<string>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<climits>
#include<complex>
#include<iostream>
#include<algorithm>
#define ll long long
using namespace std;

const int maxn = 210000;

int n,m;
vector<int>V[maxn];
struct edge{int y,nex;}a[maxn]; int len,fir[maxn];
inline void ins(const int x,const int y){a[++len]=(edge){y,fir[x]};fir[x]=len;}

int rt,d[maxn],d2[maxn];
int v[maxn],oth[maxn];
int t[maxn],tp;
void dfs(const int x)
{
	for(int k=fir[x],y=a[k].y;k;k=a[k].nex,y=a[k].y) dfs(y);
	tp=0;
	for(int k=fir[x],y=a[k].y;k;k=a[k].nex,y=a[k].y) t[++tp]=oth[y];
	for(int i=2;i<=tp;i+=2)
	{
		int x=t[i-1],y=t[i];
		V[x].push_back(y); V[y].push_back(x);
	}
	if(d[x]&1) oth[x]=t[tp];
	else oth[x]=x;
}
void col(const int x)
{
	for(int i=0;i<V[x].size();i++)
	{
		const int y=V[x][i];
		if(v[y]==2) v[y]=-v[x],col(y);
	}
}

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		int x; scanf("%d",&x);
		if(x==-1) rt=i;
		else ins(x,i),d[x]++;
	}
	dfs(rt);
	
	memset(fir,0,sizeof fir); len=0;
	for(int i=1;i<=n;i++)
	{
		int x; scanf("%d",&x);
		if(x==-1) rt=i;
		else ins(x,i),d2[x]++;
	}
	for(int i=1;i<=n;i++) if((d[i]&1)!=(d2[i]&1)) { puts("IMPOSSIBLE");return 0; }
	for(int i=1;i<=n;i++) if(!(d[i]&1)) v[i]=2;
	dfs(rt);
	
	for(int i=1;i<=n;i++) if(v[i]==2) v[i]=1,col(i);
	puts("POSSIBLE");
	for(int i=1;i<=n;i++) printf("%d ",v[i]);
	
	return 0;
}
