#include<cstdio>
#include<algorithm>
#include<cctype>
#include<vector>
#define pb push_back
#define put(x,y) ans.pb((P){x,y})
using namespace std;
#define G getchar()
int read()
{
	int x=0; char ch=G;
	for (;!isdigit(ch);ch=G);
	for (;isdigit(ch);ch=G) x=(x<<3)+(x<<1)+(ch^48);
	return x;
}
#undef G

int n; char s[260];
typedef pair<int,int> P;
vector<P> ans;
void dfs(int x,int msk,bool rev){  //from the top 1 or bottom 0
	int tot=0,pos;
	for (int i=n;i;i--) if (msk>>i-1&1) tot++,pos=i;
//	printf("dfs %d %d %d  %d %d\n",x,msk,rev,tot,pos);
	for (int i=x+1;i<pos;i++) ans.pb((P){i,1});
	if (tot==1){
		if (rev) put(pos,1),put(pos,0),put(pos-1,0);
		else put(pos-1,0),put(pos,0),put(pos,1);
	}
	else{
		if (rev){
			put(pos,1);
			dfs(pos,msk^(1<<pos-1),1);
			put(pos-1,1); put(pos-1,0); put(pos,0); put(pos,1);
			dfs(pos,msk^(1<<pos-1),0);
			put(pos,0); put(pos-1,0);
		}
		else{
			put(pos-1,0); put(pos,0); put(pos,1);
			dfs(pos,msk^(1<<pos-1),1);
			put(pos,0); put(pos-1,0); put(pos-1,1); put(pos,1);
			dfs(pos,msk^(1<<pos-1),0);
		}
	}
	for (int i=pos-1;i>=x;i--) ans.pb((P){i,1});
}
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)==j&&s[j]=='0'&&s[i]=='1') return puts("Impossible"),0;
	puts("Possible");
	ans.pb((P){0,1});
	for (int i=0;i<(1<<n);i++){
		if (s[i]=='1') continue;
		int tot=0;
		for (int j=0;j<(1<<n);j++)
			if ((i&j)==j&&s[j]=='0') tot++;
		if (tot^1) continue;
//		printf("make %d\n",i);
		dfs(0,i,1);
//		printf("size %d\n",(int)ans.size());
	}
	printf("%d\n",(int)ans.size()-1);
	for (int i=0;i<ans.size();i++) printf("%d %d\n",ans[i].first,ans[i].second);
	return 0;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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>
#include<algorithm>
#include<cctype>
#include<vector>
#define pb push_back
#define put(x,y) ans.pb((P){x,y})
using namespace std;
#define G getchar()
int read()
{
	int x=0; char ch=G;
	for (;!isdigit(ch);ch=G);
	for (;isdigit(ch);ch=G) x=(x<<3)+(x<<1)+(ch^48);
	return x;
}
#undef G

int n; char s[260];
typedef pair<int,int> P;
vector<P> ans;
void dfs(int x,int msk,bool rev){  //from the top 1 or bottom 0
	int tot=0,pos;
	for (int i=n;i;i--) if (msk>>i-1&1) tot++,pos=i;
//	printf("dfs %d %d %d  %d %d\n",x,msk,rev,tot,pos);
	for (int i=x+1;i<pos;i++) ans.pb((P){i,1});
	if (tot==1){
		if (rev) put(pos,1),put(pos,0),put(pos-1,0);
		else put(pos-1,0),put(pos,0),put(pos,1);
	}
	else{
		if (rev){
			put(pos,1);
			dfs(pos,msk^(1<<pos-1),1);
			put(pos-1,1); put(pos-1,0); put(pos,0); put(pos,1);
			dfs(pos,msk^(1<<pos-1),0);
			put(pos,0); put(pos-1,0);
		}
		else{
			put(pos-1,0); put(pos,0); put(pos,1);
			dfs(pos,msk^(1<<pos-1),1);
			put(pos,0); put(pos-1,0); put(pos-1,1); put(pos,1);
			dfs(pos,msk^(1<<pos-1),0);
		}
	}
	for (int i=pos-1;i>=x;i--) ans.pb((P){i,1});
}
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)==j&&s[j]=='0'&&s[i]=='1') return puts("Impossible"),0;
	puts("Possible");
	ans.pb((P){0,1});
	for (int i=0;i<(1<<n);i++){
		if (s[i]=='1') continue;
		int tot=0;
		for (int j=0;j<(1<<n);j++)
			if ((i&j)==j&&s[j]=='0') tot++;
		if (tot^1) continue;
//		printf("make %d\n",i);
		dfs(0,i,1);
//		printf("size %d\n",(int)ans.size());
	}
	printf("%d\n",(int)ans.size()-1);
	for (int i=0;i<ans.size();i++) printf("%d %d\n",ans[i].first,ans[i].second);
	return 0;
}
```