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 <bits/stdc++.h>

using namespace std;

#ifdef LOCAL
    #define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
    #define eprintf(...) 42
#endif

using ll = long long;
using ld = long double;
using D = double;
using uint = unsigned int;
template<typename T>
using pair2 = pair<T, T>;
using pii = pair<int, int>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second

const int maxn = 10;
const int maxmask = 1 << maxn;

char s[maxmask];
int n;
bool pin[maxn];
vector<pair2<int>> points;

int main()
{
	scanf("%d", &n);
	scanf("%s", s);
	for (int i = 0; i < n; i++) pin[i] = true;
	for (int mask = 0; mask < (1 << n); mask++) if (s[mask] == '1')
	{
		for (int i = 0; i < n; i++) if ((mask & (1 << i)))
		{
			pin[i] = false;
		}
	}
	for (int mask = 0; mask < (1 << n); mask++) if (s[mask] == '0')
	{
		bool ok = false;
		for (int i = 0; i < n; i++) if ((mask & (1 << i)))
		{
			ok |= pin[i];
		}
		if (!ok)
		{
			cout << "Impossible" << endl;
			return 0;
		}
	}
	points.pb({0, 0});
	for (int i = 1; i < n; i++) points.pb({i, 0});
	//for (int i = n; i >= 0; i--) points.pb({i, 1});
	//for (int i = 0; i < n; i++) 
	//{
		//if (pin[i])
		//{
			//points.pb({i, 0});
			//points.pb({i + 1, 0});
		//}
		//points.pb({i + 1, 1});
	//}
	for (int i = n - 1; i >= 0; i--) 
	{
		points.pb({i + 1, 0});
		if (pin[i])
		{
			points.pb({i + 1, 1});
			points.pb({i, 1});
		}
	}
	cout << "Possible" << endl;
	cout << (int)points.size() << endl;
	points.pb(points[0]);
	for (auto t : points) cout << t.fi << ' ' << t.se << endl;
    return 0;
}