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>

int n,d[200001],fa[200001],D[200001],Fa[200001],hd[400002],cnt,ans[2000001],x,y,po[200001];
bool bo[2000001];

struct node
{
	int to,next,la;
}e[2000001];

void addedge(int x,int y)
{
	e[++cnt].next=hd[x];
	e[hd[x]].la=cnt;
	hd[x]=cnt;
	e[cnt].to=y;
	e[cnt].la=0;
}

int calc(int x) {return ((x-1)^1)+1;}

void del(int x,int i)
{
	if (i==hd[x]) hd[x]=e[i].next;
	else e[e[i].la].next=e[i].next,e[e[i].next].la=e[i].la;
}

void dfs(int x)
{
	for (int i=hd[x]; i; i=e[i].next)
	{
		del(x,i); del(e[i].to,calc(i));
		ans[i]=ans[calc(i)]=i;
		dfs(e[i].to);
	}
}

int main()
{
	scanf("%d",&n); y=2*n+1;
	memset(d,0,sizeof(d));
	memset(D,0,sizeof(D));
	memset(hd,0,sizeof(hd)); cnt=0;
	for (int i=1; i<=n; i++) 
		{scanf("%d",&x),fa[i]=x; if (x==-1) addedge(y,i),addedge(i,y); else d[x]++,addedge(i,x),addedge(x,i);}
	for (int i=1; i<=n; i++) 
		{scanf("%d",&x),Fa[i]=x; if (x==-1) addedge(y,i+n),addedge(i+n,y); else D[x]++,addedge(i+n,x+n),addedge(x+n,i+n);}
	for (int i=1; i<=n; i++) if ((d[i]&1)!=(D[i]&1)) {printf("IMPOSSIBLE\n"); return 0;}
	for (int i=1; i<=n; i++) if (!(d[i]&1)) addedge(i,i+n),addedge(i+n,i),po[i]=cnt;
	dfs(y); printf("POSSIBLE\n");
	for (int i=1; i<=n; i++)
		if (d[i]&1) printf("0 "); else if (e[ans[po[i]]].to<=n) printf("-1 "); else printf("1 ");
	printf("\n");
	return 0;
}