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<bits/stdc++.h>

using namespace std;

int N, root[2], ap[100009], parity[2][100009], t[2][100009];
vector < int > h[100009], v[2][100009];

void addEdge (int x, int y)
{
    h[x].push_back (y);
    h[y].push_back (x);
}

int dfs (int lin, int nod)
{
    int curr = 0;
    for (auto it : v[lin][nod])
    {
        int curr2 = dfs (lin, it);
        if (curr != 0) addEdge (curr, curr2), curr = 0;
        else curr = curr2;
    }
    if (v[lin][nod].size () & 1)
    {
        parity[lin][nod] = 0;
        return curr;
    }
    parity[lin][nod] = 1;
    return nod;
}

void paint (int nod, int c)
{
    if (ap[nod])
    {
        assert (ap[nod] == c);
        return ;
    }
    ap[nod] = c;
    for (auto it : h[nod])
        paint (it, 3 - c);
}

int main ()
{
//freopen ("input", "r", stdin);
//freopen ("output", "w", stdout);

scanf ("%d", &N);
for (int i=0; i<2; i++)
    for (int j=1; j<=N; j++)
    {
        scanf ("%d", &t[i][j]);
        if (t[i][j] == -1) root[i] = j;
        else v[i][t[i][j]].push_back (j);
    }
dfs (0, root[0]);
dfs (1, root[1]);
for (int i=1; i<=N; i++)
    if (parity[0][i] != parity[1][i])
    {
        printf ("IMPOSSIBLE\n");
        return 0;
    }
printf ("POSSIBLE\n");
for (int i=1; i<=N; i++)
    if (ap[i] == 0 && parity[0][i])
        paint (i, 1);
for (int i=1; i<=N; i++)
    if (parity[0][i] == 0) printf ("0 ");
    else
    if (ap[i] == 1) printf ("1 ");
    else printf ("-1 ");
printf ("\n");
return 0;
}
