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<stdio.h>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 2e5+5;
struct node{
	int en[maxn],owo,la[maxn],nt[maxn];
	void adg(int x,int y) {
		en[++owo]=y; nt[owo]=la[x]; la[x] = owo;
		en[++owo]=x; nt[owo]=la[y]; la[y] = owo;
	}
}v,v1,v2;
int n;
int zb[maxn],yc[maxn],zz[maxn];
int sta[maxn],tp;
void dfs1(int x,int ba) {
	for(int it=v1.la[x];it;it=v1.nt[it]) {
		int y = v1.en[it];
		if(y==ba) continue;
		dfs1(y,x);
	}
	if(zb[x]==1) sta[++tp] = x;
	while(tp>=2) {
		v.adg(sta[tp],sta[tp-1]);
		tp-=2;
	}
}
void dfs2(int x,int ba) {
	for(int it=v2.la[x];it;it=v2.nt[it]) {
		int y = v2.en[it];
		if(y==ba) continue;
		dfs2(y,x);
	}
	if(zb[x]==1) sta[++tp] = x;
	while(tp>=2) {
		v.adg(sta[tp],sta[tp-1]);
		tp-=2;
	}
}
void dfs(int x,int vv) {
	yc[x] = vv;
	for(int it=v.la[x];it;it=v.nt[it]) {
		int y = v.en[it];
		if(yc[y]==0) dfs(y,-vv);
	}
}
int main() {
	scanf("%d",&n);
	for(int i=1;i<=n;i++) zb[i] = zz[i] = 1;
	int rt1 = 0 , rt2 = 0;
	for(int i=1;i<=n;i++) {
		int x;
		scanf("%d",&x);
		if(x==-1) rt1 = i;
		else v1.adg(x,i),zb[x]^=1;
	}
	for(int i=1;i<=n;i++) {
		int x;
		scanf("%d",&x);
		if(x==-1) rt2 = i;
		else v2.adg(x,i),zz[x]^=1;
	}
	for(int i=1;i<=n;i++) {
		if(zz[i]!=zb[i]) {
			puts("IMPOSSIBLE");
			return 0;
		}
	}
	puts("POSSIBLE");
	dfs1(rt1,0);
	tp = 0; 
	dfs2(rt2,0);
	for(int i=1;i<=n;i++) {
		if(zb[i]==1&&(yc[i]==0)) {
			dfs(i,1);
		}
	}
	for(int i=1;i<=n;i++) {
		if(zb[i]==0) printf("0 ");
		else printf("%d ",yc[i]);
	}
}