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<cstdio>
#include<vector>
#include<algorithm>
using namespace std;

int N; char str[1 << 8 | 1];
vector < int > pot;
vector < int > operator +(vector < int > p , vector < int > q){p.insert(p.end() , q.begin() , q.end()); return p;}

vector < int > build(int id , int x){
	if(id < (1 << (x + 1))) return vector < int >{x + 1 , -x - 1};
	vector < int > tmp = build(id , x + 1);
	if(!(id >> x & 1)) return vector < int >{x + 1} + tmp + vector < int >{x + 1};
	vector < int > rev = tmp; reverse(rev.begin() , rev.end());
	return vector < int >{x + 1} + tmp + vector < int >{x + 1} + vector < int >{-x - 1} + rev + vector < int >{-x - 1};
}

int main(){
	scanf("%d %s" , &N , str);
	for(int i = 0 ; i < 1 << N ; ++i)
		if(str[i] == '1')
			for(int j = 0 ; j < N ; ++j)
				if((i >> j & 1) && str[i ^ (1 << j)] != '1'){
					puts("Impossible"); return 0;
				}
	for(int i = 0 ; i < 1 << N ; ++i)
		if(str[i] == '0'){
			bool flg = 1;
			for(int j = 0 ; j < N ; ++j)
				if((i >> j & 1) && str[i ^ (1 << j)] == '0')
					flg = 0;
			if(flg) pot = pot + build(i , 0);
		}
	puts("Possible"); vector < pair < int , int > > node;
	int curx = 0 , cury = 0; node.push_back(make_pair(0 , 0));
	for(int i = 0 ; i < pot.size() ; ++i){
		if(pot[i] < 0 && cury == 1) node.push_back(make_pair(curx , cury = 0));
		if(pot[i] > 0 && cury == 0) node.push_back(make_pair(curx , cury = 1));
		int t = abs(pot[i]) - 1;
		if(curx == t) node.push_back(make_pair(++curx , cury));
		else node.push_back(make_pair(--curx , cury));
	}
	if(cury) node.push_back(make_pair(0 , 0));
	printf("%d\n" , node.size() - 1);
	for(auto t : node) printf("%d %d\n" , t.first , t.second);
	return 0;
}