#include<cstdio>
char A[1<<8|1];
int op[333333],cnt;
void make(int S,int ty){
	if(!S)return;
	int i=0;while(!(S>>i&1))i++;
	if(S==(1<<i)){
		op[cnt++]=ty&2?-i-1:i+1;
		return;
	}
	if(ty&1){
		op[cnt++]=ty&2?i+1:-i-1;
		make(S^1<<i,ty^3);
		op[cnt++]=ty&2?-i-1:i+1;
		make(S^1<<i,ty);
	}
	else{
		make(S^1<<i,ty);
		op[cnt++]=ty&2?-i-1:i+1;
		make(S^1<<i,ty^3);
		op[cnt++]=ty&2?i+1:-i-1;
	}
}
int x[333333],y[333333],L; 
int main(){
	int N;scanf("%d%s",&N,A);
	for(int S=1;S<1<<N;S++)if(A[S]=='1')
		for(int T=S;T;T=T-1&S)if(A[T]=='0')return puts("Impossible"),0;
	for(int S=1;S<1<<N;S++)if(A[S]=='0'){
		bool fl=1;
		for(int T=S-1&S;T&&fl;T=T-1&S)if(A[T]=='0')fl=0;
		if(fl)make(S,0),make(S,1);
	}
	for(int i=0;i<cnt;i++){
		int x1=op[i]>0?op[i]-1:-op[i],x2=op[i]>0?x1+1:x1-1;
		for(;x[L]!=x1;L++)x[L]<x1?x[L+1]=x[L]+1:x[L+1]=x[L]-1;
		x[++L]=x1;y[L]=1;
		x[++L]=x2;y[L]=1;
		x[++L]=x2;y[L]=0;
	}
	for(;x[L];L++)x[L+1]=x[L]-1;
	printf("Possible\n%d\n",L);
	for(int i=0;i<=L;i++)printf("%d %d\n",x[i],y[i]);
}
### 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<cstdio>
char A[1<<8|1];
int op[333333],cnt;
void make(int S,int ty){
	if(!S)return;
	int i=0;while(!(S>>i&1))i++;
	if(S==(1<<i)){
		op[cnt++]=ty&2?-i-1:i+1;
		return;
	}
	if(ty&1){
		op[cnt++]=ty&2?i+1:-i-1;
		make(S^1<<i,ty^3);
		op[cnt++]=ty&2?-i-1:i+1;
		make(S^1<<i,ty);
	}
	else{
		make(S^1<<i,ty);
		op[cnt++]=ty&2?-i-1:i+1;
		make(S^1<<i,ty^3);
		op[cnt++]=ty&2?i+1:-i-1;
	}
}
int x[333333],y[333333],L; 
int main(){
	int N;scanf("%d%s",&N,A);
	for(int S=1;S<1<<N;S++)if(A[S]=='1')
		for(int T=S;T;T=T-1&S)if(A[T]=='0')return puts("Impossible"),0;
	for(int S=1;S<1<<N;S++)if(A[S]=='0'){
		bool fl=1;
		for(int T=S-1&S;T&&fl;T=T-1&S)if(A[T]=='0')fl=0;
		if(fl)make(S,0),make(S,1);
	}
	for(int i=0;i<cnt;i++){
		int x1=op[i]>0?op[i]-1:-op[i],x2=op[i]>0?x1+1:x1-1;
		for(;x[L]!=x1;L++)x[L]<x1?x[L+1]=x[L]+1:x[L+1]=x[L]-1;
		x[++L]=x1;y[L]=1;
		x[++L]=x2;y[L]=1;
		x[++L]=x2;y[L]=0;
	}
	for(;x[L];L++)x[L+1]=x[L]-1;
	printf("Possible\n%d\n",L);
	for(int i=0;i<=L;i++)printf("%d %d\n",x[i],y[i]);
}
```