#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<string>
#include<algorithm>
using namespace std;
#define NO return !printf("Impossible\n")
#define N 433
inline int read(){
	int x=0,f=1;
	char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-')f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9'){
		x=(x<<1)+(x<<3)+c-'0';
		c=getchar();
	}
	return x*f;
}
int n,a[N];
char s[N];
vector<pair<int,int> > ans;
string Rev(string s){
	reverse(s.begin(),s.end());
	for(auto &c:s){
		if(c>='a'&&c<='z')c+='A'-'a';
		else c+='a'-'A';
	}
	return s;
}
int main(){
	n=read();
	scanf("%s",s);
	for(int i=0;i<(1<<n);++i){
		a[i]=s[i]-'0';
	}
	if(!a[0])NO;
	for(int i=0;i<(1<<n);++i){
		if(a[i]){
			for(int j=(i-1)&i;j;j=(j-1)&i){
				if(!a[j])NO;
			}
		}
	}
	ans.emplace_back(0,0);
	for(int i=0;i<(1<<n);++i){
		if(a[i])continue;
		bool ok=true;
		for(int j=(i-1)&i;j;j=(j-1)&i){
			if(!a[j]){
				ok=false;
				break;
			}
		}
		if(!ok)continue;
		string myh;
		for(int j=0;j<n;++j){
			if((i>>j)&1){
				if(myh.empty())myh=(char)('a'+j);
				else myh=(char)('a'+j)+myh+(char)('A'+j)+Rev(myh);
			}
		}
		int pos=0;
		for(auto c:myh){
			int d=0,x=0;
			if(c>='a'&&c<='z'){
				x=c-'a',d=1;
			}
			else{
				x=c-'A',d=0;
			}
			while(pos<x)ans.emplace_back(++pos,0);
			while(pos>x)ans.emplace_back(--pos,0);
			if(d){
				ans.emplace_back(x,1);
				ans.emplace_back(x+1,1);
				ans.emplace_back(x+1,0);
				ans.emplace_back(x,0);
			}
			else{
				ans.emplace_back(x+1,0);
				ans.emplace_back(x+1,1);
				ans.emplace_back(x,1);
				ans.emplace_back(x,0);
			}
		}
		while(pos>0)ans.emplace_back(--pos,0);
	}
	printf("Possible\n%d\n",(int)ans.size()-1);
	for(auto x:ans){
		printf("%d %d\n",x.first,x.second);
	}
	return 0;
}


### 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<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<string>
#include<algorithm>
using namespace std;
#define NO return !printf("Impossible\n")
#define N 433
inline int read(){
	int x=0,f=1;
	char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-')f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9'){
		x=(x<<1)+(x<<3)+c-'0';
		c=getchar();
	}
	return x*f;
}
int n,a[N];
char s[N];
vector<pair<int,int> > ans;
string Rev(string s){
	reverse(s.begin(),s.end());
	for(auto &c:s){
		if(c>='a'&&c<='z')c+='A'-'a';
		else c+='a'-'A';
	}
	return s;
}
int main(){
	n=read();
	scanf("%s",s);
	for(int i=0;i<(1<<n);++i){
		a[i]=s[i]-'0';
	}
	if(!a[0])NO;
	for(int i=0;i<(1<<n);++i){
		if(a[i]){
			for(int j=(i-1)&i;j;j=(j-1)&i){
				if(!a[j])NO;
			}
		}
	}
	ans.emplace_back(0,0);
	for(int i=0;i<(1<<n);++i){
		if(a[i])continue;
		bool ok=true;
		for(int j=(i-1)&i;j;j=(j-1)&i){
			if(!a[j]){
				ok=false;
				break;
			}
		}
		if(!ok)continue;
		string myh;
		for(int j=0;j<n;++j){
			if((i>>j)&1){
				if(myh.empty())myh=(char)('a'+j);
				else myh=(char)('a'+j)+myh+(char)('A'+j)+Rev(myh);
			}
		}
		int pos=0;
		for(auto c:myh){
			int d=0,x=0;
			if(c>='a'&&c<='z'){
				x=c-'a',d=1;
			}
			else{
				x=c-'A',d=0;
			}
			while(pos<x)ans.emplace_back(++pos,0);
			while(pos>x)ans.emplace_back(--pos,0);
			if(d){
				ans.emplace_back(x,1);
				ans.emplace_back(x+1,1);
				ans.emplace_back(x+1,0);
				ans.emplace_back(x,0);
			}
			else{
				ans.emplace_back(x+1,0);
				ans.emplace_back(x+1,1);
				ans.emplace_back(x,1);
				ans.emplace_back(x,0);
			}
		}
		while(pos>0)ans.emplace_back(--pos,0);
	}
	printf("Possible\n%d\n",(int)ans.size()-1);
	for(auto x:ans){
		printf("%d %d\n",x.first,x.second);
	}
	return 0;
}


```