Problem:
Score : 1400 points

Problem Statement

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.

Notes

C is said to be a closed curve if and only if:

C is a continuous function from [0,1] to \mathbb{R}^2 such that C(0) = C(1).

We say that a closed curve C can be continuously moved without touching a set of points X so that it becomes a closed curve D if and only if:

There exists a function f : [0,1] \times [0,1] \rightarrow \mathbb{R}^2 that satisfies all of the following.

f is continuous.

f(0,t) = C(t).

f(1,t) = D(t).

f(x,t) \notin X.

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.

Sample Input 1

1
10

Sample Output 1

Possible
4
0 0
0 1
1 1
1 0
0 0

When S = \emptyset, we can move this curve so that every point on it has a negative y-coordinate.

When S = \{0\}, we cannot do so.

Sample Input 2

2
1000

Sample Output 2

Possible
6
1 0
2 0
2 1
1 1
0 1
0 0
1 0

The output represents the following curve:

Sample Input 3

2
1001

Sample Output 3

Impossible

Sample Input 4

1
11

Sample Output 4

Possible
0
1 1

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        char[] str = sc.next().toCharArray();

        //引っかかるところを探す
        boolean a[] = new boolean[1 << n];
        for(int i = 0 ; i < (1 << n) ; i++){
            a[i] = str[i] == '0';
        }

        boolean pick[] = new boolean[n];
        for(int i = 0 ; i < n ; i++){
            pick[i] = a[1 << i];
        }

        boolean ans = true;
        for(int i = 0 ; i < (1 << n) ; i++){
            for(int j = 0 ; j < n ; j++){
                if(((i >> j)&1) == 1){
                    if(pick[j] && !a[i]){
                        ans = false;
                    }
                }
            }
        }

        System.out.println(ans?"Possible":"Impossible");
        boolean prev = false;
        int l = 0;

        if(ans){
            for(int i = 0 ; i < n ; i++){
                if(prev){
                    if(pick[i]){
                        l++;
                    }else{
                        l++;
                        l++;
                    }
                }else{
                    if(pick[i]){
                        l++;
                        l++;
                    }else{
                        l++;
                    }
                }
                prev = pick[i];
            }

            for(int i = prev?n:n-1 ; i >= 0 ; i--){
                l++;
            }

            prev = false;

            System.out.println(l);
            System.out.println("0 0");
            for(int i = 0 ; i < n ; i++){
                if(prev){
                    if(pick[i]){
                        System.out.println(""+(i+1)+" 1");
                    }else{
                        System.out.println(""+(i)+" 0");
                        System.out.println(""+(i+1)+" 0");
                    }
                }else{
                    if(pick[i]){
                        System.out.println(""+(i)+" 1");
                        System.out.println(""+(i+1)+" 1");
                    }else{
                        System.out.println(""+(i+1)+" 0");
                    }
                }
                prev = pick[i];
            }

            for(int i = (prev?n:n-1) ; i >= 0 ; i--){
                System.out.println(""+i+" 0");
            }
        }
    }
}