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<bits/stdc++.h>
using namespace std;
const int N=2e5+10;
struct E{
	int x,y,nxt,flag;
}mm[N<<2];
int h[N],len,p[N],n,ans[N],cur[N];
bool v[N];
inline int rd(){
	int x=0;int f=1;char s=getchar();
	while(!isdigit(s)) f=(s=='-'?-1:f),s=getchar();
	while(isdigit(s)) x=(x<<1)+(x<<3)+s-'0',s=getchar();
	return x*f;
}

inline void add(int x,int y){
	++len;
	mm[len].x=x;mm[len].y=y;mm[len].nxt=h[x];h[x]=len;
	++len;
	mm[len].x=y;mm[len].y=x;mm[len].nxt=h[y];h[y]=len;
	p[x]++;p[y]++;
	return ;
}
void dfs(int u,int f){
	for(int& k=h[u];k;k=mm[k].nxt){
		int y=mm[k].y;
		if(mm[k].flag||f==y) continue; 
		mm[k].flag=mm[k%2==0?k-1:k+1].flag=1;
		if(v[u] && u<=n && y>n) ans[u]=ans[y]=1; 
		if(v[u] && u>n && y<=n) ans[u]=ans[y]=-1;
		dfs(y,u);
	}
}
int main(){
//	freopen("t.in","r",stdin);
	n=rd();memset(h,0,sizeof h);
	for(int i=1;i<=n;i++){
		int u=rd();u=max(0,u);
		add(u,i);
	}
	for(int i=1;i<=n;i++){
		int u=rd();u=u==-1?0:(u+n);
		add(u,i+n);
	}
	for(int i=1;i<=2*n;i++){
		if(p[i]%2) v[i]=1;
		else v[i]=0;
	}
	bool flag=true;
	for(int i=1;i<=n;i++)
		if(v[i]!=v[i+n]) flag=0;
	if(flag) printf("POSSIBLE\n");
	else {
		printf("IMPOSSIBLE\n");
		return 0;
	}
	for(int i=1;i<=n;i++)
		if(v[i]) add(i,i+n);//,printf("#%d %d\n",i,i+n);
	dfs(1,0);
	for(int i=1;i<=n;i++) printf("%d ",ans[i]);printf("\n");
	return 0;
}