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 <cstdlib>
#include <cstring>
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <stack>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define MOD @
#define ADD(X,Y) ((X) = ((X) + (Y)%MOD) % MOD)
typedef long long i64; typedef vector<int> ivec; typedef vector<string> svec;

int N;
char A[300];
vector<pair<int, int>> ans;

/*
void draw_curve(int mask, bool flip = false)
{
	vector<pair<int, int>> seq;
	seq.push_back({0, 0});
	for (int i = 0; i < N; ++i) {
		pair<int, int> src = make_pair(i, (mask >> i) & 1);
		pair<int, int> dest = make_pair(i + 1, (mask >> i) & 1);
		if (seq.back() != src) seq.push_back(src);
		seq.push_back(dest);
	}
	if (seq.back().second == 1) seq.push_back({N, 0});
	for (int i = N - 1; i >= 0; --i) seq.push_back({i, 0});
	if (flip) reverse(seq.begin(), seq.end());
	for (int i = 1; i < seq.size(); ++i) ans.push_back(seq[i]);
}
*/
void draw_curve(int p, bool flip = false)
{
	vector<pair<int, int>> seq;
	for (int i = 0; i <= p; ++i) seq.push_back({i, 0});
	seq.push_back({p, 1});
	seq.push_back({p + 1, 1});
	for (int i = p + 1; i >= 0; --i) seq.push_back({i, 0});
	if (flip) reverse(seq.begin(), seq.end());
	for (int i = 1; i < seq.size(); ++i) ans.push_back(seq[i]);
}

int main()
{
	scanf("%d%s", &N, A);
	for (int i = 0; i < (1 << N); ++i) {
		if (A[i] == '0') {
			for (int j = 0; j < N; ++j) if ((i & (1 << j)) == 0) {
				if (A[i | (1 << j)] == '1') {
					puts("Impossible");
					return 0;
				}
			}
		}
	}
	if (A[0] == '0') {
		puts("Impossible");
		return 0;
	}
	puts("Possible");
	vector<int> minima;
	for (int i = 0; i < (1 << N); ++i) {
		if (A[i] == '0') {
			bool flg = true;
			for (int j = 0; j < N; ++j) {
				if ((i & (1 << j)) != 0 && A[i ^ (1 << j)] == '0') flg = false;
			}
			if (flg) minima.push_back(i);
		}
	}
	ans.push_back({0, 0});
	for (int m : minima) {
		vector<int> bits;
		for (int i = 0; i < N; ++i) if (m & (1 << i)) bits.push_back(i);
		vector<int> seq;
		seq.push_back(bits[0]);
		for (int i = 1; i < bits.size(); ++i) {
			vector<int> seq2;
			bool flg = false;
			for (int j = 0; j < seq.size(); ++j) {
				seq2.push_back(seq[j]);
				if (flg) seq2.push_back(~bits[i]);
				else seq2.push_back(bits[i]);
				flg = !flg;
			}
			for (int j = seq.size() - 1; j >= 0; --j) {
				seq2.push_back(~seq[j]);
				if (flg) seq2.push_back(~bits[i]);
				else seq2.push_back(bits[i]);
				flg = !flg;
			}
			seq = seq2;
		}
		for (int i = 0; i < seq.size(); ++i) {
			//if (seq[i] >= 0) printf("%d ", seq[i]);
			//else printf("!%d ", ~seq[i]);
			if (seq[i] >= 0) draw_curve(seq[i]);
			else draw_curve(~seq[i]);
		}
	}
	printf("%d\n", ans.size() - 1);
	for (auto p : ans) printf("%d %d\n", p.first, p.second);
	return 0;
}