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 <algorithm>
#include <cstring>
using namespace std;

const int MAX_N = 100005;

class Edge {
public:
  int nxt, to;
} e[MAX_N << 3];

int head[MAX_N << 1], cnt;
void addedge(int u, int v) {
  e[++cnt] = (Edge){head[u], v}, head[u] = cnt;
  e[++cnt] = (Edge){head[v], u}, head[v] = cnt;
}

int deg1[MAX_N], deg2[MAX_N];
bool vis[MAX_N << 3];
#define rev(x) (x & 1 ? x + 1 : x - 1)
		
void dfs(int u) {
  for (int &i = head[u]; i; i = e[i].nxt)
    if (!vis[i] && !vis[rev(i)]) {
      vis[i] = 1;
      dfs(e[i].to);
    }
}

int main() {
  int N;
  scanf("%d", &N);
  for (int i = 1; i <= N; ++i) {
    int fa;
    scanf("%d", &fa);
    if (fa != -1) deg1[fa] ^= 1, addedge(fa, i);
    else addedge(i, 2 * N + 1);
  }
  
  for (int i = 1; i <= N; ++i) {
    int fa;
    scanf("%d", &fa);
    if (fa != -1) deg2[fa] ^= 1, addedge(fa + N, i + N);
    else addedge(i + N, 2 * N + 1);
  }

  static int val[MAX_N];
  
  for (int i = 1; i <= N; ++i) {
    if (deg1[i] != deg2[i]) {
      puts("IMPOSSIBLE");
      return 0;
    }
    if (!deg1[i]) addedge(i, i + N), val[i] = cnt;
  }

  puts("POSSIBLE");
  dfs(1);

  for (int i = 1; i <= N; ++i)
    if (!val[i]) printf("0 ");
    else printf("%d ", vis[val[i]] ? 1 : -1);
  puts("");
  return 0;
}