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;
int main(){
  int n;
  cin>>n;
  vector<bool> a(1<<n);
  for(int i=0;i<(1<<n);i++){
    char c;
    cin>>c;
    a.at(i)=(c=='1');
  }
  vector<int> s;
  vector<bool> b(1<<n,1);
  for(int i=0;i<(1<<n);i++){
    if(a.at(i))
      continue;
    bool f=0;
    for(int x:s)
      if((x&i)==x)
        f=1;
    if(f)
      continue;
    s.push_back(i);
    for(int j=0;j<(1<<n);j++)
      b.at(j)=(b.at(j)&&((i&j)!=i));
  }
  if(a!=b){
    cout<<"Impossible"<<endl;
    return 0;
  }
  cout<<"Possible"<<endl;
  vector<vector<int>> c;
  for(int x:s){
    vector<vector<int>> cx;
    for(int i=0;i<n;i++,x/=2){
      if(x%2==0)
        continue;
      vector<vector<int>> ci;
      for(int j=0;j<n;j++){
        ci.push_back({j,0});
        if(i==j){
          ci.push_back({j,1});
          ci.push_back({j+1,1});
        }
      }
      for(int j=n;j>0;j--)
        ci.push_back({j,0});
      if(cx.empty()){
        cx=ci;
        continue;
      }
      vector<vector<int>> cxb=cx;
      vector<vector<int>> cib=ci;
      reverse(cxb.begin()+1,cxb.end());
      reverse(cib.begin()+1,cib.end());
      for(auto y:ci)
        cx.push_back(y);
      for(auto y:cxb)
        cx.push_back(y);
      for(auto y:cib)
        cx.push_back(y);
    }
    for(auto y:cx)
      c.push_back(y);
  }
  c.push_back({0,0});
  cout<<c.size()-1<<endl;
  for(auto x:c)
    cout<<x.at(0)<<' '<<x.at(1)<<endl;
}