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<cstdio>
#include<vector>
#include<algorithm>
using namespace std;

int N; char str[1 << 8];
vector < int > pot;
vector < int > operator +(vector < int > p , vector < int > q){p.insert(p.end() , q.begin() , q.end()); return p;}

vector < int > build(int id , int x){
	if(id < (1 << (x + 1))) return vector < int > {x + 1 , -x - 1};
	vector < int > tmp = build(id , x + 1);
	if(!(id >> x & 1)) return vector < int >{x + 1} + tmp + vector < int >{x + 1};
	vector < int > rev = tmp; reverse(rev.begin() , rev.end());
	return vector < int >{x + 1} + tmp + vector < int >{x + 1} + vector < int >{-x - 1} + tmp + vector < int >{-x - 1};
}

int main(){
	scanf("%d %s" , &N , str);
	for(int i = 0 ; i < 1 << N ; ++i)
		if(str[i] == '1')
			for(int j = 0 ; j < 1 << i ; ++j)
				if((i >> j & 1) && str[i ^ (1 << j)] != '1'){
					puts("Impossible"); return 0;
				}
	for(int i = 0 ; i < 1 << N ; ++i)
		if(str[i] == '0'){
			bool flg = 1;
			for(int j = 0 ; j < 1 << i ; ++j)
				if((i >> j & 1) && str[i ^ (1 << j)] == '0')
					flg = 0;
			if(flg) pot = pot + build(i , 0);
		}
	puts("Possible"); vector < pair < int , int > > node;
	int curx = 0 , cury = 0; node.push_back(make_pair(0 , 0));
	for(int i = 0 ; i < pot.size() ; ++i){
		if(pot[i] < 0 && cury == 1) node.push_back(make_pair(curx , cury = 0));
		if(pot[i] > 0 && cury == 0) node.push_back(make_pair(curx , cury = 1));
		int t = abs(pot[i]) - 1;
		if(curx == t) node.push_back(make_pair(++curx , cury));
		else node.push_back(make_pair(--curx , cury));
	}
	if(node.back() != make_pair(0 , 0)) node.push_back(make_pair(0 , 0));
	printf("%d\n" , node.size() - 1);
	for(auto t : node) printf("%d %d\n" , t.first , t.second);
	return 0;
}