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:
/*

Code for problem E by cookiedoth
Generated 21 Mar 2020 at 05.19 PM


______▄███████▄_______
______█▄█████▄█_______
______█▼▼▼▼▼█_______
_____██________ ██______
______█▲▲▲▲▲█_______
______█████████_______
_______██____ ██________

>_<
o_O
^_^

*/

#include <iostream>
#include <fstream>
#include <vector>
#include <set>
#include <map>
#include <bitset>
#include <algorithm>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <functional>
#include <unordered_set>
#include <unordered_map>
#include <string>
#include <queue>
#include <deque>
#include <stack>
#include <complex>
#include <cassert>
#include <random>
#include <cstring>
#include <numeric>
#define ll long long
#define ld long double
#define null NULL
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define debug(a) cerr << #a << " = " << a << endl
#define forn(i, n) for (int i = 0; i < n; ++i)
#define sz(a) (int)a.size()

using namespace std;

template<class T> int chkmax(T &a, T b) {
	if (b > a) {
		a = b;
		return 1;
	}
	return 0;
}

template<class T> int chkmin(T &a, T b) {
	if (b < a) {
		a = b;
		return 1;
	}
	return 0;
}

template<class iterator> void output(iterator begin, iterator end, ostream& out = cerr) {
	while (begin != end) {
		out << (*begin) << " ";
		begin++;
	}
	out << endl;
}

template<class T> void output(T x, ostream& out = cerr) {
	output(x.begin(), x.end(), out);
}

void fast_io() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
}

int x;
vector<pair<int, int> > ans;

void move(int dest_x) {
	if (x == dest_x) {
		return;
	}
	if (dest_x > x) {
		x++;
		ans.emplace_back(x, 0);
	} else {
		x--;
		ans.emplace_back(x, 0);
	}
	move(dest_x);
}

void use(int pos, int d) {
	// cerr << "use " << pos << " " << d << endl;
	if (d == 0) {
		move(pos);
		ans.emplace_back(pos, 1);
		ans.emplace_back(pos + 1, 1);
		ans.emplace_back(pos + 1, 0);
		x++;
	} else {
		move(pos + 1);
		ans.emplace_back(pos + 1, 1);
		ans.emplace_back(pos, 1);
		ans.emplace_back(pos, 0);
		x--;
	}
}

const int mx = 300;
int n, a[mx];

vector<vector<pair<int, int> > > str;

void gen_str() {
	str.resize(n + 1);
	str[1] = {{0, 0}};
	for (int i = 2; i <= n; ++i) {
		str[i].emplace_back(i - 1, 0);
		for (auto pp : str[i - 1]) {
			str[i].push_back(pp);
		}
		str[i].emplace_back(i - 1, 1);
		vector<pair<int, int> > keks = str[i - 1];
		reverse(all(keks));
		for (auto pp : keks) {
			str[i].emplace_back(pp.first, pp.second ^ 1);
		}
		// cerr << "i = " << i << endl;
		// for (auto pp : str[i]) {
		// 	cerr << pp.first << " " << pp.second << endl;
		// }
	}
}

void make(int sub) {
	vector<int> have;
	for (int i = 0; i < n; ++i) {
		if ((sub >> i) & 1) {
			have.push_back(i);
		}
	}
	int k = __builtin_popcount(sub);
	for (auto pp : str[k]) {
		use(have[pp.first], pp.second);
	}
}

signed main() {
	fast_io();
	ans.emplace_back(0, 0);
	cin >> n;
	gen_str();
	for (int i = 0; i < (1 << n); ++i) {
		char c;
		cin >> c;
		a[i] = c - '0';
	}
	if (a[0] == 0) {
		cout << "Impossible" << endl;
		exit(0);
	}
	for (int i = 0; i < (1 << n); ++i) {
		for (int j = 0; j < i; ++j) {
			if (a[i] == 1 && ((i & j) == j) && a[j] == 0) {
				cout << "Impossible" << endl;
				exit(0);
			}
		}
	}
	for (int i = 0; i < (1 << n); ++i) {
		if (a[i] == 0) {
			int ok = 1;
			for (int j = 0; j < i; ++j) {
				if (a[j] == 0 && (i & j) == j) {
					ok = 0;
					break;
				}
			}
			if (ok) {
				make(i);
			}
		}
	}
	move(0);
	cout << "Possible" << "\n";
	cout << (int)ans.size() - 1 << "\n";
	for (auto pp : ans) {
		cout << pp.first << " " << pp.second << "\n";
	}
}