#include <bits/stdc++.h>
#define fi first
#define sc second
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define all(x) x.begin(),x.end()
#define cmin(a,b) (a>b?a=b:a)
#define cmax(a,b) (a<b?a=b:a)
#define mem(a,k) memset(a,k,sizeof(a))
#define lop(i,s,t) for(int i=s;i<(t);++i)
#define rep(i,s,t) for(int i=s;i<=(t);++i)
#define dec(i,s,t) for(int i=s;i>=(t);--i)
#define fore(i,v) for(int i=g[v],d=es[i].d;i;i=es[i].nxt,d=es[i].d)
using namespace std;

#ifdef CURIOUSCAT
#define Pr(f,...) fprintf(stderr,f,##__VA_ARGS__),fflush(stderr)
#else
#define Pr(f,...) ;
#endif

typedef long long ll;

template<typename T>
void read(T &x){
	x=0; 
	char c;
	for(c=getchar();!isdigit(c);c=getchar()); 
	for(;isdigit(c);c=getchar())x=x*10+c-'0'; 
}

const int N=1e6+50;

int P;
int mul(int a,int b){ return 1ll*a*b%P; }
//int add(int a,int b){ a+=b; return a>=P?a-P:a; }
void upd(int &a,int b) { a=a+b>=P?a+b-P:a+b; }
int sub(int a,int b){ a-=b; return a<0?a+P:a; }
int gcd(int a,int b){ return !b?a:gcd(b,a%b); }


int n; string s; vector<pair<int,int>> ans;
void add(int msk,int inv){
	Pr("add %d,%d\n",msk,inv); 
	int lsb=msk&-msk;
	if(lsb==msk){
		int x=31-__builtin_clz(msk); 
		Pr("x=%d\n",x);
		rep(i,0,x)ans.eb(i,0); 
		Pr("ans.size()=%d\n",(int)ans.size()); 
		if(inv){
			ans.eb(x,1);
			ans.eb(x+1,1);
			ans.eb(x+1,0);
		}else{
			ans.eb(x+1,0);
			ans.eb(x+1,1); 
			ans.eb(x,1); 
		}
		dec(i,x,1)ans.eb(i,0);
	}else{
		if(!inv){
			add(msk^lsb,0);
			add(lsb,0);
			add(msk^lsb,1);
			add(lsb,1);
		}else{
			add(lsb,0); 
			add(msk^lsb,0);
			add(lsb,1);
			add(msk^lsb,1); 
		}
	}
}
int main(int argc,char *argv[]){
#ifdef CURIOUSCAT
#endif
	cin>>n>>s; 
	lop(i,0,(1<<n))lop(j,0,n)if(i>>j&1&&s[i]=='1'&&s[i^1<<j]=='0'){
		cout<<"Impossible"<<endl; 
		return 0; 
	}
	cout<<"Possible"<<endl;

	lop(i,0,1<<n)if(s[i]=='0')add(i,0); 
	ans.eb(0,0); 
	cout<<(int)ans.size()-1<<endl;
	for(auto x:ans)cout<<x.fi<<" "<<x.sc<<endl;
	return 0; 
}

### Prompt

In cpp, your task is to solve 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 fi first
#define sc second
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define all(x) x.begin(),x.end()
#define cmin(a,b) (a>b?a=b:a)
#define cmax(a,b) (a<b?a=b:a)
#define mem(a,k) memset(a,k,sizeof(a))
#define lop(i,s,t) for(int i=s;i<(t);++i)
#define rep(i,s,t) for(int i=s;i<=(t);++i)
#define dec(i,s,t) for(int i=s;i>=(t);--i)
#define fore(i,v) for(int i=g[v],d=es[i].d;i;i=es[i].nxt,d=es[i].d)
using namespace std;

#ifdef CURIOUSCAT
#define Pr(f,...) fprintf(stderr,f,##__VA_ARGS__),fflush(stderr)
#else
#define Pr(f,...) ;
#endif

typedef long long ll;

template<typename T>
void read(T &x){
	x=0; 
	char c;
	for(c=getchar();!isdigit(c);c=getchar()); 
	for(;isdigit(c);c=getchar())x=x*10+c-'0'; 
}

const int N=1e6+50;

int P;
int mul(int a,int b){ return 1ll*a*b%P; }
//int add(int a,int b){ a+=b; return a>=P?a-P:a; }
void upd(int &a,int b) { a=a+b>=P?a+b-P:a+b; }
int sub(int a,int b){ a-=b; return a<0?a+P:a; }
int gcd(int a,int b){ return !b?a:gcd(b,a%b); }


int n; string s; vector<pair<int,int>> ans;
void add(int msk,int inv){
	Pr("add %d,%d\n",msk,inv); 
	int lsb=msk&-msk;
	if(lsb==msk){
		int x=31-__builtin_clz(msk); 
		Pr("x=%d\n",x);
		rep(i,0,x)ans.eb(i,0); 
		Pr("ans.size()=%d\n",(int)ans.size()); 
		if(inv){
			ans.eb(x,1);
			ans.eb(x+1,1);
			ans.eb(x+1,0);
		}else{
			ans.eb(x+1,0);
			ans.eb(x+1,1); 
			ans.eb(x,1); 
		}
		dec(i,x,1)ans.eb(i,0);
	}else{
		if(!inv){
			add(msk^lsb,0);
			add(lsb,0);
			add(msk^lsb,1);
			add(lsb,1);
		}else{
			add(lsb,0); 
			add(msk^lsb,0);
			add(lsb,1);
			add(msk^lsb,1); 
		}
	}
}
int main(int argc,char *argv[]){
#ifdef CURIOUSCAT
#endif
	cin>>n>>s; 
	lop(i,0,(1<<n))lop(j,0,n)if(i>>j&1&&s[i]=='1'&&s[i^1<<j]=='0'){
		cout<<"Impossible"<<endl; 
		return 0; 
	}
	cout<<"Possible"<<endl;

	lop(i,0,1<<n)if(s[i]=='0')add(i,0); 
	ans.eb(0,0); 
	cout<<(int)ans.size()-1<<endl;
	for(auto x:ans)cout<<x.fi<<" "<<x.sc<<endl;
	return 0; 
}

```