#include<bits/stdc++.h>
using namespace std;
string A;
inline void Exit(){
	puts("Impossible");
	exit(0);
}
int n;
typedef pair<int,int> pi;
vector<int> ths;
vector<int> operator + (vector<int> a, vector<int> b){
	for(size_t i=0;i<b.size();i++){
		a.push_back(b[i]);
	}return a;
}
inline vector<int> Rev(vector<int> a){
	reverse(a.begin(),a.end());
	for(size_t i=0;i<a.size();i++)a[i]=-a[i];
	return a;
}

vector<pi> pit;

int main(){
	cin >> n >> A;
	if(A[0]=='0')Exit();
	for(int i=0;i<1<<n;i++){
		if(A[i]=='1')for(int j=i;j;j=i&(j-1)){
			if(A[j]=='0')Exit();
		}
	}
	for(int s=0;s<1<<n;s++){
		int flg=0;
		if(A[s]=='0'){
			for(int j=(s-1)&s;j;j=s&(j-1)){
				if(A[j]=='0')flg=1;
			}
			if(flg)continue;
			ths.clear();
			for(int i=0;i<n;i++){
				if(s&(1<<i)){
					if(!ths.size())ths = ths + vector<int>(1,i+1);
					else ths = vector<int>(1,i+1) + ths + vector<int>(1,-(i+1)) + Rev(ths);
				}
			}
			for(size_t i=0;i<ths.size();i++){
				int j = ths[i];
				int d = abs(j)-1;
				for(int i=0;i<=d;i++)pit.push_back(pi(i,0));
				if(j<0){
					pit.push_back(pi(d,1));
					pit.push_back(pi(d+1,1));
					pit.push_back(pi(d+1,0));
				}else{
					pit.push_back(pi(d+1,0));
					pit.push_back(pi(d+1,1));
					pit.push_back(pi(d,1));
				}
				for(int i=d;i;i--)pit.push_back(pi(i,0));
			}
		}
	}
	pit.push_back(pi(0,0));
	cout << "Possible\n" << pit.size()-1 << endl;
	for(size_t i=0;i<pit.size();i++){
		printf("%d %d\n",pit[i].first,pit[i].second);
	}
}

### Prompt

Your challenge is to write a cpp solution to the following 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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
string A;
inline void Exit(){
	puts("Impossible");
	exit(0);
}
int n;
typedef pair<int,int> pi;
vector<int> ths;
vector<int> operator + (vector<int> a, vector<int> b){
	for(size_t i=0;i<b.size();i++){
		a.push_back(b[i]);
	}return a;
}
inline vector<int> Rev(vector<int> a){
	reverse(a.begin(),a.end());
	for(size_t i=0;i<a.size();i++)a[i]=-a[i];
	return a;
}

vector<pi> pit;

int main(){
	cin >> n >> A;
	if(A[0]=='0')Exit();
	for(int i=0;i<1<<n;i++){
		if(A[i]=='1')for(int j=i;j;j=i&(j-1)){
			if(A[j]=='0')Exit();
		}
	}
	for(int s=0;s<1<<n;s++){
		int flg=0;
		if(A[s]=='0'){
			for(int j=(s-1)&s;j;j=s&(j-1)){
				if(A[j]=='0')flg=1;
			}
			if(flg)continue;
			ths.clear();
			for(int i=0;i<n;i++){
				if(s&(1<<i)){
					if(!ths.size())ths = ths + vector<int>(1,i+1);
					else ths = vector<int>(1,i+1) + ths + vector<int>(1,-(i+1)) + Rev(ths);
				}
			}
			for(size_t i=0;i<ths.size();i++){
				int j = ths[i];
				int d = abs(j)-1;
				for(int i=0;i<=d;i++)pit.push_back(pi(i,0));
				if(j<0){
					pit.push_back(pi(d,1));
					pit.push_back(pi(d+1,1));
					pit.push_back(pi(d+1,0));
				}else{
					pit.push_back(pi(d+1,0));
					pit.push_back(pi(d+1,1));
					pit.push_back(pi(d,1));
				}
				for(int i=d;i;i--)pit.push_back(pi(i,0));
			}
		}
	}
	pit.push_back(pi(0,0));
	cout << "Possible\n" << pit.size()-1 << endl;
	for(size_t i=0;i<pit.size();i++){
		printf("%d %d\n",pit[i].first,pit[i].second);
	}
}

```