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;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> pii;
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define sz(a) int(a.size())
const int N=5e5+10;

int gi() {
	int x=0,o=1;char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
	if(ch=='-') o=-1,ch=getchar();
	while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
	return x*o;
}

int to[N<<1],nxt[N<<1],tt=1,n,rt1,rt2,h[N],id[N];
vi E1[N],E2[N];
bool vis[N],dir[N];

void adde(int u,int v) {
	to[++tt]=v;nxt[tt]=h[u];h[u]=tt;
	to[++tt]=u;nxt[tt]=h[v];h[v]=tt;
}

void dfs(int u) {
	for(int &i=h[u],v;v=to[i],i;i=nxt[i]) {
		if(vis[i>>1]) continue;
		dir[i>>1]=i&1;vis[i>>1]=1;dfs(v);
	}
}

int main() {
	cin>>n;
	for(int i=1;i<=n;i++) {
		int p=gi();
		if(p<0) rt1=i; else E1[p].pb(i);
	}
	for(int i=1;i<=n;i++) {
		int p=gi();
		if(p<0) rt2=i; else E2[p].pb(i);
	}
	for(int i=1;i<=n;i++) if((sz(E1[i])&1)!=(sz(E2[i])&1)) return cout<<"IMPOSSIBLE",0;
	cout<<"POSSIBLE\n";
	for(int i=1;i<=n;i++) {
		for(auto j:E1[i]) adde(i,j);
		for(auto j:E2[i]) adde(i+n,j+n);
		if(sz(E1[i])&1^1) adde(i,i+n),id[i]=tt>>1;
	}
	adde(0,rt1);adde(0,rt2+n);
	dfs(0);
	for(int i=1;i<=n;i++) {
		if(sz(E1[i])&1) cout<<0<<' ';
		else if(dir[id[i]]) cout<<1<<' ';
		else cout<<-1<<' ';
	}
	return 0;
}