#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
#include <bits/stdc++.h>
using namespace std;
template<class t> inline t read(t &x){
	char c=getchar();bool f=0;x=0;
	while(!isdigit(c)) f|=c=='-',c=getchar();
	while(isdigit(c)) x=(x<<1)+(x<<3)+(c^48),c=getchar();
	if(f) x=-x;return x;
}
template<class t,class ...A> inline void read(t &x,A &...a){
	read(x);read(a...);
}
template<class t> inline void write(t x){
	if(x<0) putchar('-'),write(-x);
	else{if(x>9) write(x/10);putchar('0'+x%10);}
}

#define pii pair<int,int>
#define x first
#define y second

const int N=405;
int n,a[N];
char s[N];
vector<pii> ans;

string rev(string s){
	reverse(s.begin(),s.end());
	for(char &c:s){
		if(c>='a') c-='a'-'A';
		else c+='a'-'A';
	}
	return s;
}

signed main(){
	read(n);
	for(int i=0;i<(1<<n);i++) scanf("%1d",&a[i]);
	if(!a[0]) return puts("Impossible"),0;
	for(int i=0;i<(1<<n);i++) if(a[i])
		for(int j=(i-1)&i;j;j=(j-1)&i) if(!a[j])
			return puts("Impossible"),0;
	ans.emplace_back(0,0);
	for(int i=0;i<(1<<n);i++) if(!a[i]){
		bool flag=0;
		for(int j=(i-1)&i;j;j=(j-1)&i) if(!a[j]){
			flag=1;
			break;
		}
		if(flag) continue;
		string s;
		for(int j=0;j<n;j++) if(i>>j&1){
			if(s.empty()) s=(char)('a'+j);
			else s=(char)('a'+j)+s+(char)('A'+j)+rev(s);
		}
		int pos=0;
		for(char c:s){
			int x=0,d=0;
			if(c>='a') x=c-'a',d=1;
			else x=c-'A',d=0;
			while(pos<x) ans.emplace_back(++pos,0);
			while(pos>x) ans.emplace_back(--pos,0);
			if(d){
				ans.emplace_back(x,1);
				ans.emplace_back(x+1,1);
				ans.emplace_back(x+1,0);
				ans.emplace_back(x,0);
			}
			else{
				ans.emplace_back(x+1,0);
				ans.emplace_back(x+1,1);
				ans.emplace_back(x,1);
				ans.emplace_back(x,0);
			}
		}
		while(pos) ans.emplace_back(--pos,0);
	}
	puts("Possible");
	write(ans.size()-1);puts("");
	for(auto x:ans) write(x.x),putchar(' '),write(x.y),puts("");
}
### Prompt

Generate a Cpp solution to 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
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
#include <bits/stdc++.h>
using namespace std;
template<class t> inline t read(t &x){
	char c=getchar();bool f=0;x=0;
	while(!isdigit(c)) f|=c=='-',c=getchar();
	while(isdigit(c)) x=(x<<1)+(x<<3)+(c^48),c=getchar();
	if(f) x=-x;return x;
}
template<class t,class ...A> inline void read(t &x,A &...a){
	read(x);read(a...);
}
template<class t> inline void write(t x){
	if(x<0) putchar('-'),write(-x);
	else{if(x>9) write(x/10);putchar('0'+x%10);}
}

#define pii pair<int,int>
#define x first
#define y second

const int N=405;
int n,a[N];
char s[N];
vector<pii> ans;

string rev(string s){
	reverse(s.begin(),s.end());
	for(char &c:s){
		if(c>='a') c-='a'-'A';
		else c+='a'-'A';
	}
	return s;
}

signed main(){
	read(n);
	for(int i=0;i<(1<<n);i++) scanf("%1d",&a[i]);
	if(!a[0]) return puts("Impossible"),0;
	for(int i=0;i<(1<<n);i++) if(a[i])
		for(int j=(i-1)&i;j;j=(j-1)&i) if(!a[j])
			return puts("Impossible"),0;
	ans.emplace_back(0,0);
	for(int i=0;i<(1<<n);i++) if(!a[i]){
		bool flag=0;
		for(int j=(i-1)&i;j;j=(j-1)&i) if(!a[j]){
			flag=1;
			break;
		}
		if(flag) continue;
		string s;
		for(int j=0;j<n;j++) if(i>>j&1){
			if(s.empty()) s=(char)('a'+j);
			else s=(char)('a'+j)+s+(char)('A'+j)+rev(s);
		}
		int pos=0;
		for(char c:s){
			int x=0,d=0;
			if(c>='a') x=c-'a',d=1;
			else x=c-'A',d=0;
			while(pos<x) ans.emplace_back(++pos,0);
			while(pos>x) ans.emplace_back(--pos,0);
			if(d){
				ans.emplace_back(x,1);
				ans.emplace_back(x+1,1);
				ans.emplace_back(x+1,0);
				ans.emplace_back(x,0);
			}
			else{
				ans.emplace_back(x+1,0);
				ans.emplace_back(x+1,1);
				ans.emplace_back(x,1);
				ans.emplace_back(x,0);
			}
		}
		while(pos) ans.emplace_back(--pos,0);
	}
	puts("Possible");
	write(ans.size()-1);puts("");
	for(auto x:ans) write(x.x),putchar(' '),write(x.y),puts("");
}
```