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>
#include <ext/pb_ds/assoc_container.hpp>
#include <decimal/decimal>
using namespace std;
using namespace __gnu_pbds;

#define rep(i, s, t) for (int i = s; i < int(t); ++i)

#define sc	scanf
#define pr	printf
#define ff	first
#define ss	second
#define mp	make_pair
#define pb	push_back
#define eb	emplace_back
#define range(c)	c.begin(), c.end()
#define lowbit(x)	((x) & -(x))
#define pop_count	__builtin_popcount
#define pop_count_64	__builtin_popcountll
#define tail_count	__builtin_ctz
#define tail_count_64	__builtin_ctzll
#define lead_count	__builtin_clz
#define lead_count_64	__builtin_clzll

#define TEMP template<typename
#define CONST(type) constexpr type const

TEMP... Ts> struct is_con { static CONST(bool) value = false; };
TEMP... Ts> struct is_con<vector<Ts...>> { static CONST(bool) value = true; };
TEMP... Ts> struct is_con<list<Ts...>> { static CONST(bool) value = true; };
TEMP... Ts> struct is_con<deque<Ts...>> { static CONST(bool) value = true; };
TEMP... Ts> struct is_con<set<Ts...>> { static CONST(bool) value = true; };
TEMP... Ts> struct is_con<unordered_set<Ts...>> { static CONST(bool) value = true; };
TEMP... Ts> struct is_con<map<Ts...>> { static CONST(bool) value = true; };
TEMP... Ts> struct is_con<unordered_map<Ts...>> { static CONST(bool) value = true; };

TEMP T1, typename T2> ostream& operator<<(ostream& os, pair<T1, T2> const& p) { return os << "<" << p.ff << "," << p.ss << ">"; }
TEMP T> typename enable_if<is_con<T>::value, ostream>::type& operator<<(ostream& os, T const& c) {
	os << "{";
	bool first = true;
	for (auto const& x : c) {
		if (first) first = false;
		else os << ",";
		os << " " << x;
	}
	return os << " }";
}

TEMP T, typename... Ts> void dbg(T h, Ts... t) { cerr << " " << h; dbg<Ts...>(t...); }
template<> void dbg<int>(int x) { cerr << endl; }

#ifdef LOCAL
#define debug(...) cerr << "[ "#__VA_ARGS__" ] :", dbg(__VA_ARGS__, 0)
#else
#define debug(...) 0
#endif

TEMP T, typename U> using ordered_map = tree<T, U, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
TEMP T> using ordered_set = ordered_map<T, null_type>;

typedef long long i64;
typedef decimal::decimal128 i99;
typedef unsigned int u32;
typedef unsigned long long u64;
typedef float f32;
typedef double f64;
typedef long double f80;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<i64> vl;

TEMP T> T& minn(T& lhs, T const& rhs) { return lhs = min(lhs, rhs); }
TEMP T> T& maxx(T& lhs, T const& rhs) { return lhs = max(lhs, rhs); }

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int n;
	string a;
	cin >> n >> a;
	vi dp(1 << n);
	rep(s, 0, 1 << n) dp[s] = a[s] == '0';
	debug(dp);
	rep(i, 0, n) rep(s, 0, 1 << n) if (s >> i & 1) dp[s] |= dp[s ^ 1 << i];
	debug(dp);
	bool ok = true;
	rep(s, 0, 1 << n) ok &= dp[s] ^ (a[s] == '1');
	vi clique;
	rep(s, 0, 1 << n) if (dp[s]) {
		bool check = true;;
		rep(t, 0, 1 << n) if (t != s && (s & t) == t) check &= !dp[t];
		if (check) clique.pb(s);
	}
	int tot = 0, cnt = 0;
	for (int s : clique) {
		int x = pop_count(s);
		ok &= x <= 2;
		tot |= s;
	}
	if (!ok) return cout << "Impossible" << endl, 0;
	cout << "Possible" << endl;
	debug(clique);
	vector<pii> ans(1);
	for (int s : clique) {
		if (s == lowbit(s)) {
			int x = tail_count(s);
			rep(i, 1, x + 1) ans.eb(i, 0);
			ans.eb(x + 1, 0);
			ans.eb(x + 1, 1);
			ans.eb(x, 1);
			for (int i = x; i >= 0; --i) ans.eb(i, 0);
		} else {
			int x = tail_count(s);
			s ^= lowbit(s);
			int y = tail_count(s);
			rep(i, 1, y + 1) ans.eb(i, 0);
			ans.eb(y + 1, 0);
			ans.eb(y + 1, 1);
			ans.eb(y, 1);
			for (int i = y; i >= x; --i) ans.eb(i, 0);
			rep(i, x, y + 2) ans.eb(i, 1);
			for (int i = y + 1; i > x; --i) ans.eb(i, 0);
			ans.eb(x + 1, 1);
			ans.eb(x, 1);
			for (int i = x; i >= 0; --i) ans.eb(i, 0);
		}
	}
	cout << ans.size() - 1 << endl;
	for (auto const& p : ans) cout << p.ff << " " << p.ss << endl;
}