#include<bits/stdc++.h>
using namespace std;
char s[300];int n;
struct Node {int x,y;Node(){}Node(int _x,int _y){x=_x;y=_y;}};
vector<Node> V;
inline string inv(string s)
{
	reverse(s.begin(),s.end());
	for(int i=0;i<(int)s.size();i++)
	{
		if(s[i]>='A'&&s[i]<='Z') s[i]+='a'-'A';
		else s[i]-='a'-'A';
	}
	return s;
}
int main()
{
	scanf("%d%s",&n,s);
	for(int i=0;i<(1<<n);i++) for(int j=0;j<(1<<n);j++)
		if((i&j)==(s[i]=='1'?j:i)&&s[j]!=s[i]) return puts("Impossible"),0;
	puts("Possible");
	V.push_back(Node(0,0));
	for(int i=0;i<(1<<n);i++) if(s[i]=='0')
	{
		bool flag=1;
		for(int j=0;j<i;j++) if((i&j)==j&&s[j]=='0'){flag=0;break;}
		if(!flag) continue;
		string S;
		for(int j=0;j<n;j++) if((i>>j)&1)
		{
			if(S.size()==0)S+='a'+j;
			else S=(char)('a'+j)+S+(char)('A'+j)+inv(S);
		}
		int lstv=0;
		for(int j=0;j<(int)S.size();j++)
		{
			int now,fv;
			if(S[j]>='A'&&S[j]<='Z') now=S[j]-'A',fv=0;
			else now=S[j]-'a',fv=1;
			for(int k=lstv+1;k<=now;k++) V.push_back(Node(k,0));
			for(int k=lstv-1;k>=now;k--) V.push_back(Node(k,0));
			if(fv)
			{
				V.push_back(Node(now,1));V.push_back(Node(now+1,1));
				V.push_back(Node(now+1,0));V.push_back(Node(now,0));
			}
			else
			{
				V.push_back(Node(now+1,0));V.push_back(Node(now+1,1));
				V.push_back(Node(now,1));V.push_back(Node(now,0));
			}
			lstv=now;
		}
		for(int j=lstv-1;j>=0;j--) V.push_back(Node(j,0));
	}
	cout<<V.size()-1<<endl;
	for(int i=0;i<(int)V.size();i++) printf("%d %d\n",V[i].x,V[i].y);
	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<bits/stdc++.h>
using namespace std;
char s[300];int n;
struct Node {int x,y;Node(){}Node(int _x,int _y){x=_x;y=_y;}};
vector<Node> V;
inline string inv(string s)
{
	reverse(s.begin(),s.end());
	for(int i=0;i<(int)s.size();i++)
	{
		if(s[i]>='A'&&s[i]<='Z') s[i]+='a'-'A';
		else s[i]-='a'-'A';
	}
	return s;
}
int main()
{
	scanf("%d%s",&n,s);
	for(int i=0;i<(1<<n);i++) for(int j=0;j<(1<<n);j++)
		if((i&j)==(s[i]=='1'?j:i)&&s[j]!=s[i]) return puts("Impossible"),0;
	puts("Possible");
	V.push_back(Node(0,0));
	for(int i=0;i<(1<<n);i++) if(s[i]=='0')
	{
		bool flag=1;
		for(int j=0;j<i;j++) if((i&j)==j&&s[j]=='0'){flag=0;break;}
		if(!flag) continue;
		string S;
		for(int j=0;j<n;j++) if((i>>j)&1)
		{
			if(S.size()==0)S+='a'+j;
			else S=(char)('a'+j)+S+(char)('A'+j)+inv(S);
		}
		int lstv=0;
		for(int j=0;j<(int)S.size();j++)
		{
			int now,fv;
			if(S[j]>='A'&&S[j]<='Z') now=S[j]-'A',fv=0;
			else now=S[j]-'a',fv=1;
			for(int k=lstv+1;k<=now;k++) V.push_back(Node(k,0));
			for(int k=lstv-1;k>=now;k--) V.push_back(Node(k,0));
			if(fv)
			{
				V.push_back(Node(now,1));V.push_back(Node(now+1,1));
				V.push_back(Node(now+1,0));V.push_back(Node(now,0));
			}
			else
			{
				V.push_back(Node(now+1,0));V.push_back(Node(now+1,1));
				V.push_back(Node(now,1));V.push_back(Node(now,0));
			}
			lstv=now;
		}
		for(int j=lstv-1;j>=0;j--) V.push_back(Node(j,0));
	}
	cout<<V.size()-1<<endl;
	for(int i=0;i<(int)V.size();i++) printf("%d %d\n",V[i].x,V[i].y);
	return 0;
}
```