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<iostream>
#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
const int N=1<<10;
int n;
char s[N];
int p[N];
vector<pair<int,int> > solve(int S,int now)
{
	vector<pair<int,int> >res;
	for(int i=now+1;i<n;i++)
		if(S&(1<<i))
		{
			vector<pair<int,int> > t=solve(S,i);
			for(int j=now;j<=i-1;j++)
				res.push_back(make_pair(j,1));
			for(auto [x,y]:t)
				res.push_back(make_pair(x,y));
			for(int j=i;j>=now;j--)
				res.push_back(make_pair(j,1));
			res.push_back(make_pair(now,0));
			res.push_back(make_pair(now+1,0));
			for(int j=now+1;j<=i;j++)
				res.push_back(make_pair(j,1));
			reverse(t.begin(),t.end());
			for(auto [x,y]:t)
				res.push_back(make_pair(x,y));
			for(int j=i-1;j>=now+1;j--)
				res.push_back(make_pair(j,1));
			res.push_back(make_pair(now+1,0));
			res.push_back(make_pair(now,0));
			return res;
		}
	res.push_back(make_pair(now,1));
	res.push_back(make_pair(now+1,1)); 
	res.push_back(make_pair(now+1,0));
	res.push_back(make_pair(now,0));
	return res;
}
int main()
{
	scanf("%d",&n);
	scanf("%s",s);
	for(int i=0;i<(1<<n);i++)
		p[i]=s[i]-'0';
	if(p[0]==0)
	{
		printf("Impossible");
		return 0;
	}
	for(int s=0;s<(1<<n);s++)
		if(p[s]==1)
			for(int i=s;i;i=(i-1)&s)
				if(p[i]==0)
				{
					printf("Impossible");
					return 0;
				}
	vector<pair<int,int> >ans;
	ans.push_back(make_pair(0,0));
	for(int s=1;s<(1<<n);s++)
		if(p[s]==0)
		{
			bool flag=true;
			for(int i=(s-1)&s;i;i=(i-1)&s)
				if(p[i]==0)
				{
					flag=false;
					break;
				}
			if(!flag) continue;
			vector<pair<int,int> >res;
			vector<pair<int,int> >t;
			int u=0;
			for(u=0;u<n;u++)
				if(s&(1<<u))
				{
					t=solve(s,u);
					break;
				}
			if(ans.back()!=make_pair(0,0)) res.push_back(make_pair(0,0));
			for(int i=0;i<=u-1;i++)
				res.push_back(make_pair(i,1));
			for(auto [x,y]:t)
				res.push_back(make_pair(x,y));
			for(int i=u;i>=0;i--)
				res.push_back(make_pair(i,1));
			res.push_back(make_pair(0,0));
			for(auto [x,y]:res)
				ans.push_back(make_pair(x,y));
		}
	printf("Possible\n");
	int len=ans.size()-1;
	printf("%d\n",len);
	for(auto [x,y]:ans)
		printf("%d %d\n",x,y);
	return 0;
}