Problem: 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.

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.

Examples

Input

1
10


Output

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


Input

2
1000


Output

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


Input

2
1001


Output

Impossible


Input

1
11


Output

Possible
0
1 1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

#define itn int
#define all(x) (x).begin(), (x).end()
#define make_unique(x) sort(all((x))); (x).resize(unique(all((x))) - (x).begin())

using namespace std;

inline int nxt() {
	int x;
	scanf("%d", &x);
	return x;
}

int main() {
	int n = nxt();
	string s;
	cin >> s;
	if (s == string(1 << n, '1')) {
		cout << "Possible\n";
		cout << "0\n0 0\n";
		return 0;
	}
	if (s[0] == '0') {
		cout << "Impossible\n";
		return 0;
	}
	const int N = 1 << n;
	vector<char> base(N);
	for (int i = 0; i < N; ++i) {
		if (s[i] == '0') {
			base[i] = 1;
			for (int j = i; j > 0; j = (j - 1) & i) {
				if (s[i ^ j] == '0') {
					base[i] = 0;
				}
			}
			continue;
		}
		for (int j = i; j > 0; j = (j - 1) & i) {
			if (s[i ^ j] == '0') {
				cout << "Impossible\n";
				return 0;
			}
		}
	}

	vector<int> ans;
	function<void(int, bool, bool)> add = [&](int mask, bool rev, bool flip) {
		assert(mask);
		if (__builtin_popcount(mask) == 1) {
			ans.push_back(flip ? -__builtin_ctz(mask) - 1 : __builtin_ctz(mask) + 1);
			return;
		}
		int nm = mask ^ (1 << (__builtin_ctz(mask)));
		if (rev) {
			ans.push_back(flip ? -__builtin_ctz(mask) - 1 : __builtin_ctz(mask) + 1);
			add(nm, false, !flip);
			ans.push_back(!flip ? -__builtin_ctz(mask) - 1 : __builtin_ctz(mask) + 1);
			add(nm, true, flip);
		} else {
			add(nm, false, flip);
			ans.push_back(!flip ? -__builtin_ctz(mask) - 1 : __builtin_ctz(mask) + 1);
			add(nm, true, !flip);
			ans.push_back(flip ? -__builtin_ctz(mask) - 1 : __builtin_ctz(mask) + 1);
		}
	};
	for (int mask = 0; mask < N; ++mask) {
		if (base[mask]) {
			add(mask, false, false);
		}
	}
	int y = 0;
	vector<pair<int, int>> answer = {{0, y}};
	auto go = [&](int c) {
		while (y < c) {
			y += 1;
			answer.push_back({0, y});
		}
		while (y > c) {
			y -= 1;
			answer.push_back({0, y});
		}
	};

	for (int i : ans) {
		// cerr << i << " ";
		if (i < 0) {
			go(-i - 1);
			answer.push_back({1, y});
			answer.push_back({1, y + 1});
			answer.push_back({0, y + 1});
			y += 1;
		} else {
			go(i);
			answer.push_back({1, y});
			answer.push_back({1, y - 1});
			answer.push_back({0, y - 1});
			y -= 1;
		}
	}
	// cerr << "\n";
	go(0);
	cout << "Possible\n";
	cout << (int)answer.size() - 1 << "\n";
	for (const auto& p : answer) {
		cout << p.second << " " << p.first << "\n";
	}

	return 0;
}