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<bits/stdc++.h>
using namespace std;

typedef long long LL;

const int N=8;

struct point{
  int x,y;
  point(int X=0,int Y=0){x=X;y=Y;}
};

int n;
char a[(1<<N)+9];

void into(){
  scanf("%d%s",&n,a);
}

bool Check_a(){
  for (int u=1;u<1<<n;++u){
	if (a[u]=='0') continue;
	for (int v=u-1&u;v;v=v-1&u)
	  if (a[v]=='0') return 0;
  }
  return 1;
}

vector<point>Dfs_ans(int s,int last){
  int p=0;
  for (;s>>p&1^1;++p);
  vector<point>res;
  if (s==1<<p){
	for (int i=last;i<p;++i) res.push_back(point(i,1));
	res.push_back(point(p,1));
	res.push_back(point(p,0));
	for (int i=p-1;i>=last;--i) res.push_back(point(i,1));
	return res;
  }
  vector<point>t=Dfs_ans(s^1<<p,p+1);
  for (int i=last;i<p;++i) res.push_back(point(i,1));
  res.push_back(point(p,1));
  for (int vs=t.size(),i=0;i<vs;++i) res.push_back(t[i]);
  res.push_back(point(p,1));
  res.push_back(point(p,0));
  for (int vs=t.size(),i=vs-1;i>=0;--i) res.push_back(t[i]);
  res.push_back(point(p,0));
  for (int i=p-1;i>=last;--i) res.push_back(point(i,1));
  return res;
}

int ans0;
vector<point>ans;

void Get_ans(){
  if (!(ans0=Check_a())) return;
  vector<point>tmp;
  for (int u=1;u<1<<n;++u){
	if (a[u]=='1') continue;
	int flag=1;
	for (int v=u-1&u;v;v=v-1&u)
      if (a[v]=='0') {flag=0;break;}
    if (!flag) continue;
	vector<point>t=Dfs_ans(u,0);
	for (int vs=t.size(),i=0;i<vs;++i) tmp.push_back(t[i]);
  }
  int x=0,y=0;
  ans.push_back(point(0,0));
  for (int vs=tmp.size(),i=0;i<vs;++i){
	if (y^tmp[i].y) ans.push_back(point(x,y=tmp[i].y));
	ans.push_back(point(x==tmp[i].x?++x:--x,y));
  }
  if (y) ans.push_back(point(x,y=0));
}

void work(){
  Get_ans();
}

void outo(){
  puts(ans0?"Possible":"Impossible");
  if (!ans0) return;
  int vs=ans.size();
  printf("%d\n",vs-1);
  for (int i=0;i<vs;++i)
	printf("%d %d\n",ans[i].x,ans[i].y);
}

int main(){
  into();
  work();
  outo();
  return 0;
}