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 <stdio.h>
#include <iostream>
#include <algorithm>
#include <memory.h>

using namespace std;
typedef long long LL;
const int maxn = 800005;

int ehead[maxn],ecnt=1,deg[maxn];
struct edge {
	int u,v,next;
}edg[maxn];
void add(int u,int v) {
	edg[++ecnt]=(edge){u,v,ehead[u]};
	ehead[u]=ecnt;deg[u]++;
	edg[++ecnt]=(edge){v,u,ehead[v]};
	ehead[v]=ecnt;deg[v]++;
}
int n,a[maxn],b[maxn],w[maxn],vis[maxn],ra,rb;

void Euler(int u)
{
	for (int &j=ehead[u];j;j=edg[j].next)
	if (!vis[j>>1]) {
		vis[j>>1]=true;
		int v=edg[j].v;Euler(v);
		if (u==v+n) w[v]=-1;
		if (v==u+n) w[u]=1;
	}
}
int main()
{
	#ifdef Amberframe
		freopen("agc018f.in","r",stdin);
		freopen("agc018f.out","w",stdout);
	#endif
	scanf("%d",&n);
	for (int i=1;i<=n;i++) scanf("%d",&a[i]);
	for (int i=1;i<=n;i++) scanf("%d",&b[i]);
	for (int i=1;i<=n;i++) if (a[i]!=-1) add(i,a[i]);else ra=i;
	for (int i=1;i<=n;i++) if (b[i]!=-1) add(i+n,b[i]+n);else rb=i+n;
	add(0,ra);add(0,rb);
	for (int i=1;i<=n;i++)
		if ((deg[i]&1)!=(deg[i+n]&1)) {puts("IMPOSSIBLE");return 0;}
	puts("POSSIBLE");
	for (int i=1;i<=n;i++) if (deg[i]&1) add(i,i+n);
	Euler(0);for (int i=1;i<=n;i++) printf("%d ",w[i]);
	return 0;
}