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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define BT (1<<9)
using namespace std;
typedef pair <int,int> P;

char A[BT];
string ans;
int n;

int popcount(int S)
{
	int cnt=0;
	for(int i=0;i<n;i++) if(S>>i&1) cnt++;
	return cnt;
}
string get(int S)
{
	int cnt=popcount(S);
	if(cnt==2)
	{
		vector <int> v;
		for(int i=0;i<n;i++) if(S>>i&1) v.push_back(i);
		string ret="";
		ret+='0'+v[0];
		ret+='0'+v[1];
		ret+='0'+v[0]+n;
		ret+='0'+v[1]+n;
		return ret;
	}
	for(int i=0;i<n;i++)
	{
		if(S>>i&1)
		{
			string A=get(S-(1<<i));
			string ret="";
			ret+='0'+i;
			ret+=A;
			ret+='0'+i+n;
			for(int j=(int) (A.size())-1;j>=0;j--)
			{
				int y=A[j]-'0';
				if(y>=n) y-=n;
				else y+=n;
				ret+='0'+y;
			}
			return ret;
		}
	}
	return "";
}
void print(string ret)
{
	int x=0;
	vector <P> ans;
	for(int i=0;i<ret.size();i++)
	{
		int y=ret[i]-'0';
		if(y<n)
		{
			while(x<y)
			{
				ans.push_back(P(x,0));
				x++;
			}
			while(x>y)
			{
				ans.push_back(P(x,0));
				x--;
			}
			ans.push_back(P(x,0));
			ans.push_back(P(x,1));
			x++;
			ans.push_back(P(x,1));
		}
		else
		{
			y-=n;
			while(x<=y)
			{
				ans.push_back(P(x,0));
				x++;
			}
			while(x>y+1)
			{
				ans.push_back(P(x,0));
				x--;
			}
			ans.push_back(P(x,0));
			ans.push_back(P(x,1));
			x--;
			ans.push_back(P(x,1));
		}
	}
	while(x>=0)
	{
		ans.push_back(P(x,0));
		x--;
	}
	puts("Possible");
	printf("%d\n",ans.size()-1);
	for(int i=0;i<ans.size();i++) printf("%d %d\n",ans[i].first,ans[i].second);
}
int main()
{
	scanf("%d",&n);
	scanf("%s",&A);
	for(int S=0;S<(1<<n);S++)
	{
		for(int T=0;T<(1<<n);T++)
		{
			if((T|S)==S&&A[S]=='1'&&A[T]=='0')
			{
				puts("Impossible");
				return 0;
			}
		}
	}
	ans="";
	for(int S=0;S<(1<<n);S++)
	{
		if(A[S]=='0')
		{
			bool up=true;
			for(int i=0;i<n;i++)
			{
				if(S>>i&1)
				{
					if(A[S-(1<<i)]=='0')
					{
						up=false;
					}
				}
			}
			if(!up) continue;
			if(popcount(S)>=2) ans+=get(S);
			else
			{
				int x;
				for(int i=0;i<n;i++) if(S>>i&1) x=i;
				ans+='0'+x;
			}
			//printf("%d %s\n",S,get(S).c_str());
		}
	}
	//printf("%s\n",ans.c_str());
	print(ans);
	return 0;
}