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;
#define rep(it,st,en) for(int it=(st);it<(int)(en);++it)
#define allof(c) (c).begin(), (c).end()
#define mt make_tuple
#define mp make_pair
#define pb push_back
#define X first
#define Y second
typedef long long int ll; 
typedef long double ld;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> pii;
const int INF=(int)1e9; 
const double EPS=(ld)1e-7;

int n;

vvi bip;

int dfs(int x, int p, vvi& g){
    vi Q;
    int ret = x;
    for(int y:g[x]) if(y != p)
        Q.pb(dfs(y,x,g));
    g[x] = move(Q);
    if(g[x].size()%2){
        ret = g[x].back();
    }
    return ret;
}

vi vis;

void solve(vvi& g, int r){
    dfs(r,-1,g);
    rep(i,0,n){
        rep(j,0,g[i].size()/2){
            bip[g[i][2*j+0]].pb(g[i][2*j+1]);
            bip[g[i][2*j+1]].pb(g[i][2*j+0]);
        }
    }
}

void d(int x, int col = 1){
    if(vis[x]){
        assert(vis[x] == col);
        return;
    }
    vis[x] = col;
    for(int y:bip[x]) d(y,-col);
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cin>>n;
    vi a(n),b(n);
    vvi ga(n), gb(n);
    int ra = -1, rb = -1;
    rep(i,0,n){ cin>>a[i]; --a[i]; if(a[i]>=0) ga[a[i]].pb(i); else ra = i; }
    rep(i,0,n){ cin>>b[i]; --b[i]; if(b[i]>=0) gb[b[i]].pb(i); else rb = i; }
    rep(i,0,n){
        if(ga[i].size()%2 != gb[i].size()%2){
            cout<<"IMPOSSIBLE"<<endl;
            return 0;
        }
    }
    bip.resize(n);
    vis.resize(n,0);
    solve(ga,ra);
    solve(gb,rb);
    rep(i,0,n) if(!vis[i]) d(i);

    cout<<"POSSIBLE"<<endl;
    rep(i,0,n){
        if(ga[i].size()%2==1) cout<<0<<" ";
        else cout<<vis[i]<<" ";
    }
    cout<<endl;
    return 0;
}