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 fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

int read() {
	char ch;int sig=1;
	for(ch=getchar();ch<'0'||ch>'9';ch=getchar()) if (ch=='-') sig=-1;
	int x=ch-'0';
	for(ch=getchar();ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0';
	return x*sig;
}

const int N=2e5+5,M=1e6+5;

int t[M],nxt[M],lst[N],l;

void add(int x,int y) {
	t[++l]=y;nxt[l]=lst[x];lst[x]=l;
	t[++l]=x;nxt[l]=lst[y];lst[y]=l;
}

int n,rt1,rt2,S;
bool odd[N];

void dfs(int x,int y) {
	odd[x]=1;
	for(int i=lst[x];i;i=nxt[i])
		if (t[i]!=y) {
			dfs(t[i],x);
			odd[x]^=1;
		}
}

int b[M],an[N],top;
bool vis[M];


void find(int x) {
	for(int i=lst[x];i;i=lst[x]) {
		lst[x]=nxt[i];
		if (!vis[i]) {
			vis[i]=vis[i^1]=1;
			find(t[i]);
		}
	}
	b[++top]=x;
}

int main() {
	n=read();l=1;
	fo(i,1,n) {
		int x=read();
		if (x==-1) rt1=i;
		else add(x,i);
	}
	fo(i,n+1,n*2) {
		int x=read();
		if (x==-1) rt2=i;
		else add(x+n,i);
	}
	dfs(rt1,0);dfs(rt2,0);
	bool ok=1;
	fo(i,1,n)
		if (odd[i]^odd[i+n]) {
			ok=0;
			break;
		}
	if (!ok) {
		puts("IMPOSSIBLE");
		return 0;
	}
	puts("POSSIBLE");
	S=n*2+1;
	add(S,rt1);add(S,rt2);
	fo(i,1,n) if (odd[i]) add(i,i+n);
	find(S);
	fo(i,1,n) if (odd[i]) an[i]=-1;
	fo(i,1,top-1) if (b[i]<=n&&b[i+1]==b[i]+n) an[b[i]]=1;
	fo(i,1,n) printf("%d ",an[i]); 
	return 0;
}