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>
#define ll long long
using namespace std;
const int N=1000010,P=998244353;
inline int read(){
	int x=0,f=0,c=getchar();
	for(;c>'9'||c<'0';f=c=='-',c=getchar());
	for(;c>='0'&&c<='9';c=getchar())
	x=(x<<1)+(x<<3)+c-'0';return f?-x:x;
}
inline void write(ll x){
	if(x<0)putchar('-'),x=-x;
	if(x>9)write(x/10);putchar(x%10+'0');
}
int L=1,h[N],ne[N],to[N],f[N],sz[N],n,g[N];
void addl(int x,int y){
	ne[++L]=h[x];h[x]=L;to[L]=y;
	ne[++L]=h[y];h[y]=L;to[L]=x;
}
int tp,st[N];
void dfs(int x){
	int k;
	for(k=h[x],h[x]=ne[h[x]];k;k=h[x],h[x]=ne[h[x]])
	if(!f[k/2])f[k/2]=k%2+1,dfs(to[k]);st[++tp]=x;
}
int main(){
	//freopen("1.out","w",stdout);
	int i;
	n=read();
	for(i=1;i<=n+n;i++){
		int x=read();
		if(x<0)x=n+n+1;
		else if(i>n)x+=n;
		sz[x]++;addl(x,i);
	}
	for(i=1;i<=n;i++)if(sz[i]%2!=sz[i+n]%2){
		puts("IMPOSSIBLE");return 0;
	}else{if(sz[i]%2==0)addl(i,n+i),g[i]=L/2;}
	dfs(n+n+1);
	for(i=1;i<=n;i++)
	if(sz[i]%2==0){
		if(f[g[i]]==2)g[i]=-1;else g[i]=1;
		g[i+n]=-g[i];
	}
	puts("POSSIBLE");
	for(i=1;i<=n;i++)printf("%d ",g[i]);
	return 0;
}