Problem:
Score : 1400 points

Problem Statement

Given are a positive integer N and a sequence of length 2^N consisting of 0s and 1s: A_0,A_1,\ldots,A_{2^N-1}.
Determine whether there exists a closed curve C that satisfies the condition below for all 2^N sets S \subseteq \{0,1,\ldots,N-1 \}. If the answer is yes, construct one such closed curve.

Let x = \sum_{i \in S} 2^i and B_S be the set of points \{ (i+0.5,0.5) | i \in S \}.

If there is a way to continuously move the closed curve C without touching B_S so that every point on the closed curve has a negative y-coordinate, A_x = 1.

If there is no such way, A_x = 0.

For instruction on printing a closed curve, see Output below.

Notes

C is said to be a closed curve if and only if:

C is a continuous function from [0,1] to \mathbb{R}^2 such that C(0) = C(1).

We say that a closed curve C can be continuously moved without touching a set of points X so that it becomes a closed curve D if and only if:

There exists a function f : [0,1] \times [0,1] \rightarrow \mathbb{R}^2 that satisfies all of the following.

f is continuous.

f(0,t) = C(t).

f(1,t) = D(t).

f(x,t) \notin X.

Constraints

1 \leq N \leq 8

A_i = 0,1 \quad (0 \leq i \leq 2^N-1)

A_0 = 1

Input

Input is given from Standard Input in the following format:

N
A_0A_1 \cdots A_{2^N-1}

Output

If there is no closed curve that satisfies the condition, print Impossible.

If such a closed curve exists, print Possible in the first line.
Then, print one such curve in the following format:

L
x_0 y_0
x_1 y_1
:
x_L y_L

This represents the closed polyline that passes (x_0,y_0),(x_1,y_1),\ldots,(x_L,y_L) in this order.

Here, all of the following must be satisfied:

0 \leq x_i \leq N, 0 \leq y_i \leq 1, and x_i, y_i are integers. (0 \leq i \leq L)

|x_i-x_{i+1}| + |y_i-y_{i+1}| = 1. (0 \leq i \leq L-1)

(x_0,y_0) = (x_L,y_L).

Additionally, the length of the closed curve L must satisfy 0 \leq L \leq 250000.

It can be proved that, if there is a closed curve that satisfies the condition in Problem Statement, there is also a closed curve that can be expressed in this format.

Sample Input 1

1
10

Sample Output 1

Possible
4
0 0
0 1
1 1
1 0
0 0

When S = \emptyset, we can move this curve so that every point on it has a negative y-coordinate.

When S = \{0\}, we cannot do so.

Sample Input 2

2
1000

Sample Output 2

Possible
6
1 0
2 0
2 1
1 1
0 1
0 0
1 0

The output represents the following curve:

Sample Input 3

2
1001

Sample Output 3

Impossible

Sample Input 4

1
11

Sample Output 4

Possible
0
1 1

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cassert>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<queue>
#include<stack>
using namespace std;
#define MOD 1000000007
#define MOD2 998244353
#define INF ((1<<30)-1)
#define LINF (1LL<<60)
#define EPS (1e-10)
typedef long long Int;
typedef pair<Int, Int> P;

void ok(){
    cout << "Yes" << endl;
    exit(0);
}

void ng(){
    cout <<  "Impossible" << endl;
    exit(0);
}


int n;
int a[1 << 8];

vector<int> loops;

void entangle(int x){
    vector<int> loop;
    for(int i = n-1;i >= 0;i--){
        if((x >> i) % 2 == 0)continue;
        if(loop.empty()){
            loop.push_back(i+1);
            loop.push_back(-i-1);
        }
        else{
            vector<int> tmp;
            tmp.push_back(i+1);
            tmp.push_back(-i-1);
            for(auto it = loop.begin();it != loop.end();it++)tmp.push_back(*it);
            tmp.push_back(i+1);
            for(auto it = loop.rbegin();it != loop.rend();it++)tmp.push_back(*it);
            loop = tmp;
        }
    }
    loop.push_back(INF);
    for(auto elem:loop)loops.push_back(elem);
}

void show(){
    cout << "Possible" << endl;
    vector<P> ps;
    ps.push_back(P(0,0));
    int nowx = 0, nowy = 0;
    for(int x:loops){
        if(x == INF){
            while(nowx > 0){
                ps.push_back(P(--nowx, nowy));
            }
            continue;
        }
        if(x > 0){
            while(nowx > x){
                ps.push_back(P(--nowx, nowy));
            }
            while(nowx < x-1){
                ps.push_back(P(++nowx, nowy));
            }
            ps.push_back(P(nowx,++nowy));
            if(nowx == x)ps.push_back(P(--nowx, nowy));
            else ps.push_back(P(++nowx, nowy));
            ps.push_back(P(nowx,--nowy));
        }
        if(x < 0){
            x = x * -1;
            while(nowx > x){
                ps.push_back(P(--nowx, nowy));
            }
            while(nowx < x-1){
                ps.push_back(P(++nowx, nowy));
            }
            if(nowx == x)ps.push_back(P(--nowx, nowy));
            else ps.push_back(P(++nowx, nowy));
        }
    }
    cout << ps.size() - 1 << endl;
    for(auto p:ps){
        cout << p.first << " " << p.second << endl;
    }
}

int main(){
    cin >> n;
    for(int i = 0;i < (1<<n);i++){
        scanf("%1d", a+i);
        bool bad;
        if(a[i] == 0)bad = true;
        for(int j = 0;j < i;j++){
            if((j & i) != j)continue;
            if(a[i] == 1 && a[j] == 0){
                ng();                    
            }
            if(a[j] == 0)bad = false;
        }
        if(bad){
            entangle(i);
        }
    }
    show();
    return 0;
}