#include<bits/stdc++.h>
#define ll long long
#define re register
#define cs const

using std::cerr;
using std::cout;
using pii=std::pair<int,int>;
#define fi first
#define se second

std::string inv(std::string s){
	std::string res="";
	std::reverse(s.begin(),s.end());
	for(char c:s)
		if(isupper(c))res+=c-'A'+'a';
		else res+=c-'a'+'A';
	return res;
}

int n;
std::string A;

void Main(){
	std::cin>>n>>A;
	if(A[0]=='0'){
		puts("Impossible");
		return ;
	}for(int re i=0;i<(1<<n);++i)
		if(A[i]=='1')for(int re j=i;j;j=i&(j-1))
			if(A[j]=='0'){
				puts("Impossible");
				return ;
			}
	std::vector<pii> p({{0,0}});
	for(int re s=0;s<(1<<n);++s)if(A[s]=='0'){
		bool flag=true;
		for(int re t=s&(s-1);t&&flag;t=s&(t-1))
			if(A[t]=='0')flag=false;
		if(!flag)continue;
		std::string q;
		for(int re i=0;i<n;++i)
			if(s&(1<<i)){
				if(q.empty())q+='a'+i;
				else q=(char)('a'+i)+q+(char)('A'+i)+inv(q);
			}
		int nw=0;
		for(char c:q){
			int i,dir;
			if(isupper(c))
				i=c-'A',dir=0;
			else 
				i=c-'a',dir=1;
			for(int j=nw+1;j<=i;++j)
				p.push_back({j,0});
			for(int j=nw-1;j>=i;--j)
				p.push_back({j,0});
			nw=i;
			if(dir){
				p.push_back({i,1});
				p.push_back({i+1,1});
				p.push_back({i+1,0});
				p.push_back({i,0});
			}else {
				p.push_back({i+1,0});
				p.push_back({i+1,1});
				p.push_back({i,1});
				p.push_back({i,0});
			}
		}
		for(int j=nw-1;~j;--j)
			p.push_back({j,0});
	}
	cout<<"Possible\n"<<(p.size()-1)<<"\n";
	for(auto t:p)cout<<t.fi<<" "<<t.se<<"\n";
}

inline void file(){
#ifdef zxyoi
	freopen("E.in","r",stdin);
#endif
}signed main(){file();Main();return 0;}


### Prompt

Your task is to create 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>
#define ll long long
#define re register
#define cs const

using std::cerr;
using std::cout;
using pii=std::pair<int,int>;
#define fi first
#define se second

std::string inv(std::string s){
	std::string res="";
	std::reverse(s.begin(),s.end());
	for(char c:s)
		if(isupper(c))res+=c-'A'+'a';
		else res+=c-'a'+'A';
	return res;
}

int n;
std::string A;

void Main(){
	std::cin>>n>>A;
	if(A[0]=='0'){
		puts("Impossible");
		return ;
	}for(int re i=0;i<(1<<n);++i)
		if(A[i]=='1')for(int re j=i;j;j=i&(j-1))
			if(A[j]=='0'){
				puts("Impossible");
				return ;
			}
	std::vector<pii> p({{0,0}});
	for(int re s=0;s<(1<<n);++s)if(A[s]=='0'){
		bool flag=true;
		for(int re t=s&(s-1);t&&flag;t=s&(t-1))
			if(A[t]=='0')flag=false;
		if(!flag)continue;
		std::string q;
		for(int re i=0;i<n;++i)
			if(s&(1<<i)){
				if(q.empty())q+='a'+i;
				else q=(char)('a'+i)+q+(char)('A'+i)+inv(q);
			}
		int nw=0;
		for(char c:q){
			int i,dir;
			if(isupper(c))
				i=c-'A',dir=0;
			else 
				i=c-'a',dir=1;
			for(int j=nw+1;j<=i;++j)
				p.push_back({j,0});
			for(int j=nw-1;j>=i;--j)
				p.push_back({j,0});
			nw=i;
			if(dir){
				p.push_back({i,1});
				p.push_back({i+1,1});
				p.push_back({i+1,0});
				p.push_back({i,0});
			}else {
				p.push_back({i+1,0});
				p.push_back({i+1,1});
				p.push_back({i,1});
				p.push_back({i,0});
			}
		}
		for(int j=nw-1;~j;--j)
			p.push_back({j,0});
	}
	cout<<"Possible\n"<<(p.size()-1)<<"\n";
	for(auto t:p)cout<<t.fi<<" "<<t.se<<"\n";
}

inline void file(){
#ifdef zxyoi
	freopen("E.in","r",stdin);
#endif
}signed main(){file();Main();return 0;}


```