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>
#define LL long long
#define db long double

using namespace std;
const int N=(1<<20)+10;
int rd()
{
    int x=0,w=1;char ch=0;
    while(ch<'0'||ch>'9'){if(ch=='-') w=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+(ch^48);ch=getchar();}
    return x*w;
}
int n,f1[N],f2[N],z1[N],z2[N],an[N];
vector<int> e1[N],e2[N],ee[N];
bool vs[N];
int d1(int x)
{
    int nn=e1[x].size(),an=!z1[x]?x:0;
    for(int i=0;i<nn;++i)
    {
	int y=e1[x][i],w=d1(y);
	if(!an||!w) an|=w;
	else
	{
	    ee[an].push_back(w),ee[w].push_back(an);
	    an=0;
	}
    }
    return an;
}
int d2(int x)
{
    int nn=e2[x].size(),an=!z2[x]?x:0;
    for(int i=0;i<nn;++i)
    {
	int y=e2[x][i],w=d2(y);
	if(!an||!w) an|=w;
	else
	{
	    ee[an].push_back(w),ee[w].push_back(an);
	    an=0;
	}
    }
    return an;
}

int main()
{
    n=rd();
    for(int i=1;i<=n;++i)
    {
	f1[i]=rd();
	if(~f1[i]) e1[f1[i]].push_back(i),z1[f1[i]]^=1;
    }
    for(int i=1;i<=n;++i)
    {
	f2[i]=rd();
	if(~f2[i]) e2[f2[i]].push_back(i),z2[f2[i]]^=1;
    }
    for(int i=1;i<=n;++i)
	if(z1[i]^z2[i]){puts("IMPOSSIBLE");return 0;}
    puts("POSSIBLE");
    for(int i=1;i<=n;++i)
	if(f1[i]<0){d1(i);break;}
    for(int i=1;i<=n;++i)
	if(f2[i]<0){d2(i);break;}
    queue<int> q;
    for(int i=1;i<=n;++i)
	if(!z1[i]&&!vs[i])
	{
	    an[i]=-1,vs[i]=1,q.push(i);
	    while(!q.empty())
	    {
		int x=q.front();
		q.pop();
		int nn=ee[x].size();
		for(int j=0;j<nn;++j)
		{
		    int y=ee[x][j];
		    if(!vs[y]) vs[y]=1,an[y]=-an[x],q.push(y);
		}
	    }
	}
    for(int i=1;i<=n;++i) printf("%d ",an[i]);
    return 0;
}