#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
int n;
char p[260];
struct point {
	int x, ck, dir;
};
vector<point>Res;
vector<point> Add(int x, int mask) {
	vector <point> r, T;
	if (mask == 1) {
		r.push_back({ x,0,0 });
		r.push_back({ x,1,1 });
		return r;
	}
	if (!(mask&1)) {
		r.push_back({ x,0,0 });
		T = Add(x + 1, mask >> 1);
		r.insert(r.end(),T.begin(),T.end());
		r.push_back({ x,0,1 });
		return r;
	}
	r.push_back({ x,0,0 });
	T = Add(x + 1, mask >> 1);
	r.insert(r.end(), T.begin(), T.end());
	r.push_back({ x,0,1 });
	reverse(T.begin(), T.end());
	for (int i = 0; i < T.size(); i++)T[i].dir = !T[i].dir;
	r.push_back({ x,1,0 });
	r.insert(r.end(), T.begin(), T.end());
	r.push_back({ x,1,1 });
	return r;
}
struct PP {
	int x, y;
};
vector<PP>RR;
void Print(point a) {
	PP t1 = { a.x + a.dir, a.ck };
	PP t2 = { a.x + 1 - a.dir, a.ck };
	if (!RR.empty() && RR.back().x == t1.x && RR.back().y == t2.y) {
		RR.push_back(t2);
	}
	else {
		RR.push_back(t1);
		RR.push_back(t2);
	}
}
int main() {
	int i, j;
	scanf("%d", &n);
	scanf("%s", p);
	int s = 0;
	for (i = 0; i < (1 << n); i++) {
		int c = 0;
		if (p[i] == '0')c++, s++;
		for (j = 0; j < i; j++) {
			if ((i&j) == j) {
				if (p[j] == '0' && p[i] == '1') {
					puts("Impossible");
					return 0;
				}
				if (p[j] == '0')c--;
			}
		}
		if (c == 1) {
			vector<point>T = Add(0, i);
			Res.insert(Res.end(), T.begin(), T.end());
		}
	}
	puts("Possible");
	if (!s) {
		printf("0\n0 0\n");
		return 0;
	}
	for (auto &t : Res) {
		Print(t);
	}
	if (RR.back().x == RR[0].x && RR.back().y == RR[0].y)RR.pop_back();
	printf("%d\n", RR.size());
	RR.push_back(RR[0]);
	for(auto &t :RR)printf("%d %d\n", t.x, t.y);
}
### Prompt

Construct a CPP code solution to the problem outlined:

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
#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
int n;
char p[260];
struct point {
	int x, ck, dir;
};
vector<point>Res;
vector<point> Add(int x, int mask) {
	vector <point> r, T;
	if (mask == 1) {
		r.push_back({ x,0,0 });
		r.push_back({ x,1,1 });
		return r;
	}
	if (!(mask&1)) {
		r.push_back({ x,0,0 });
		T = Add(x + 1, mask >> 1);
		r.insert(r.end(),T.begin(),T.end());
		r.push_back({ x,0,1 });
		return r;
	}
	r.push_back({ x,0,0 });
	T = Add(x + 1, mask >> 1);
	r.insert(r.end(), T.begin(), T.end());
	r.push_back({ x,0,1 });
	reverse(T.begin(), T.end());
	for (int i = 0; i < T.size(); i++)T[i].dir = !T[i].dir;
	r.push_back({ x,1,0 });
	r.insert(r.end(), T.begin(), T.end());
	r.push_back({ x,1,1 });
	return r;
}
struct PP {
	int x, y;
};
vector<PP>RR;
void Print(point a) {
	PP t1 = { a.x + a.dir, a.ck };
	PP t2 = { a.x + 1 - a.dir, a.ck };
	if (!RR.empty() && RR.back().x == t1.x && RR.back().y == t2.y) {
		RR.push_back(t2);
	}
	else {
		RR.push_back(t1);
		RR.push_back(t2);
	}
}
int main() {
	int i, j;
	scanf("%d", &n);
	scanf("%s", p);
	int s = 0;
	for (i = 0; i < (1 << n); i++) {
		int c = 0;
		if (p[i] == '0')c++, s++;
		for (j = 0; j < i; j++) {
			if ((i&j) == j) {
				if (p[j] == '0' && p[i] == '1') {
					puts("Impossible");
					return 0;
				}
				if (p[j] == '0')c--;
			}
		}
		if (c == 1) {
			vector<point>T = Add(0, i);
			Res.insert(Res.end(), T.begin(), T.end());
		}
	}
	puts("Possible");
	if (!s) {
		printf("0\n0 0\n");
		return 0;
	}
	for (auto &t : Res) {
		Print(t);
	}
	if (RR.back().x == RR[0].x && RR.back().y == RR[0].y)RR.pop_back();
	printf("%d\n", RR.size());
	RR.push_back(RR[0]);
	for(auto &t :RR)printf("%d %d\n", t.x, t.y);
}
```