#pragma comment(linker, "/STACK:1000000000")
#pragma GCC optimize("Ofast,no-stack-protector")
#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <algorithm>
#include <memory.h>
#include <cassert>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <cmath>
#include <ctime>
#include <random>
#include <bitset>
#include <limits.h>
#include <fstream>

#define mp make_pair
#define all(a) (a).begin(), (a).end()
#define pll pair <ll, ll>

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef double ld;
#define uint unsigned int

const int maxlog = 8;
const int maxn = (1 << 8);

int a[maxn + 10];
char s[maxn + 10];

vector <pii> ans;

void go(int &x, int &y, int dx, int dy) {
	x += dx;
	y += dy;
	ans.push_back(mp(x, y));
}

vector <pii> build(vector <int> p) {
	if ((int)p.size() == 1) {
		return { mp(p[0], 1) };
	}

	int x = p.back();
	p.pop_back();

	vector <pii> ans = build(p);
	vector <pii> res;

	for (int i = 0; i < (int)ans.size(); i++) {
		res.push_back(mp(ans[i].first, -ans[i].second));
	}

	reverse(all(res));
	ans.insert(ans.end(), all(res));
	res.clear();

	for (int i = 0; i < (int)ans.size(); i++) {
		res.push_back(mp(x, (i % 2 == 0 ? 1 : -1)));
		res.push_back(ans[i]);
	}

	return res;
}

void make(int &x, int &y, pii f) {
	int ps = f.first;
	if (f.second == -1) {
		ps++;
	}

	while (x < ps) {
		x++;
		ans.push_back(mp(x, y));
	}

	while (x > ps) {
		x--;
		ans.push_back(mp(x, y));
	}

	y--;
	ans.push_back(mp(x, y));
	x += f.second;
	ans.push_back(mp(x, y));
	y++;
	ans.push_back(mp(x, y));
}

int main() {
	int n;

	cin >> n;
	cin >> s;

	for (int i = 0; i < (1 << n); i++) {
		a[i] = s[i] - '0';
	}

	for (int i = (1 << n) - 1; i >= 0; i--) {
		if (a[i] == 0) {
			for (int j = 0; j < n; j++) {
				if ((i >> j) & 1) {

				}
				else {
					if (a[i | (1 << j)] == 1) {
						cout << "Impossible\n";
						return 0;
					}
				}
			}
		}
	}

	if (a[0] == 0) {
		cout << "Impossible\n";
		return 0;
	}

	ans.push_back(mp(0, 0));

	for (int i = 0; i < (1 << n); i++) {
		if (a[i] == 1) {
			continue;
		}

		vector <int> p;

		bool nd = true;

		for (int j = 0; j < n; j++) {
			if ((i >> j) & 1) {
				if (a[i ^ (1 << j)] == 0) {
					nd = false;
				}
				p.push_back(j);
			}
		}

		if (!nd) {
			continue;
		}

		int x = 0, y = 0;

		vector <pii> to = build(p);

		y++;
		ans.push_back(mp(x, y));

		for (pii f : to) {
			make(x, y, f);
		}

		while (x > 0) {
			x--;
			ans.push_back(mp(x, y));
		}

		y--;
		ans.push_back(mp(x, y));
	}

	cout << "Possible\n";
	cout << (int)ans.size() - 1 << '\n';

	for (pii f : ans) {
		cout << f.first << ' ' << f.second << '\n';
	}
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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

### Response

```cpp
#pragma comment(linker, "/STACK:1000000000")
#pragma GCC optimize("Ofast,no-stack-protector")
#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <algorithm>
#include <memory.h>
#include <cassert>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <cmath>
#include <ctime>
#include <random>
#include <bitset>
#include <limits.h>
#include <fstream>

#define mp make_pair
#define all(a) (a).begin(), (a).end()
#define pll pair <ll, ll>

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef double ld;
#define uint unsigned int

const int maxlog = 8;
const int maxn = (1 << 8);

int a[maxn + 10];
char s[maxn + 10];

vector <pii> ans;

void go(int &x, int &y, int dx, int dy) {
	x += dx;
	y += dy;
	ans.push_back(mp(x, y));
}

vector <pii> build(vector <int> p) {
	if ((int)p.size() == 1) {
		return { mp(p[0], 1) };
	}

	int x = p.back();
	p.pop_back();

	vector <pii> ans = build(p);
	vector <pii> res;

	for (int i = 0; i < (int)ans.size(); i++) {
		res.push_back(mp(ans[i].first, -ans[i].second));
	}

	reverse(all(res));
	ans.insert(ans.end(), all(res));
	res.clear();

	for (int i = 0; i < (int)ans.size(); i++) {
		res.push_back(mp(x, (i % 2 == 0 ? 1 : -1)));
		res.push_back(ans[i]);
	}

	return res;
}

void make(int &x, int &y, pii f) {
	int ps = f.first;
	if (f.second == -1) {
		ps++;
	}

	while (x < ps) {
		x++;
		ans.push_back(mp(x, y));
	}

	while (x > ps) {
		x--;
		ans.push_back(mp(x, y));
	}

	y--;
	ans.push_back(mp(x, y));
	x += f.second;
	ans.push_back(mp(x, y));
	y++;
	ans.push_back(mp(x, y));
}

int main() {
	int n;

	cin >> n;
	cin >> s;

	for (int i = 0; i < (1 << n); i++) {
		a[i] = s[i] - '0';
	}

	for (int i = (1 << n) - 1; i >= 0; i--) {
		if (a[i] == 0) {
			for (int j = 0; j < n; j++) {
				if ((i >> j) & 1) {

				}
				else {
					if (a[i | (1 << j)] == 1) {
						cout << "Impossible\n";
						return 0;
					}
				}
			}
		}
	}

	if (a[0] == 0) {
		cout << "Impossible\n";
		return 0;
	}

	ans.push_back(mp(0, 0));

	for (int i = 0; i < (1 << n); i++) {
		if (a[i] == 1) {
			continue;
		}

		vector <int> p;

		bool nd = true;

		for (int j = 0; j < n; j++) {
			if ((i >> j) & 1) {
				if (a[i ^ (1 << j)] == 0) {
					nd = false;
				}
				p.push_back(j);
			}
		}

		if (!nd) {
			continue;
		}

		int x = 0, y = 0;

		vector <pii> to = build(p);

		y++;
		ans.push_back(mp(x, y));

		for (pii f : to) {
			make(x, y, f);
		}

		while (x > 0) {
			x--;
			ans.push_back(mp(x, y));
		}

		y--;
		ans.push_back(mp(x, y));
	}

	cout << "Possible\n";
	cout << (int)ans.size() - 1 << '\n';

	for (pii f : ans) {
		cout << f.first << ' ' << f.second << '\n';
	}
}
```