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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

import java.io.BufferedReader
import java.io.InputStream
import java.io.InputStreamReader
import java.io.PrintWriter
import java.lang.StringBuilder
import java.util.*

fun PrintWriter.solve(sc: FastScanner) {
    val n = sc.nextInt()
    val a = sc.nextLine().toCharArray().map { it.toString().toInt() }
    val contain = Array(n) { false }
    var ok = true
    for (i in 0 until n) {
        val j = Math.pow(2.0, i.toDouble()).toInt()
        if (a[j] == 0) {
            contain[i] = true
        }
    }
    val l = a.count()
    for (i in 0 until l) {
        var iszero = false
        for (j in 0 until n) {
            if ((i and 1.shl(j)) != 0 && contain[j]) {
                iszero = true
                break
            }
        }
        if ((iszero && a[i] != 0) || (!iszero && a[i] == 0)) {
            ok = false
            break
        }
    }
    var change = 0
    for (i in 1 until n) {
        if (contain[i] != contain[i-1]) change++
    }
    if (!contain[0]) change--
    if (change >= 2) ok = false
    if (ok) {
        println("Possible")
        if (change < 0) {
            println(0)
            println("1 1")
            return
        }
        val points = mutableListOf<Pair<Int,Int>>()
        var up = false
        points.add(0 to 0)
        for (i in 0 until n) {
            if (contain[i]) {
                if (!up) {
                    up = true
                    points.add(i to 1)
                }
                points.add(i + 1 to 1)
            } else {
                if (up) {
                    up = false
                    points.add(i to 0)
                }
                points.add(i + 1 to 0)
            }
        }
        if (up) {
            points.add(n to 0)
        }
        for (i in n - 1 downTo 0) {
            points.add(i to 0)
        }
        println(points.count() - 1)
        for (p in points) {
            println("%d %d".format(p.first, p.second))
        }
    } else {
        println("Impossible")
    }
}

fun main(args: Array<String>) {
    val writer = PrintWriter(System.out, false)
    writer.solve(FastScanner(System.`in`))
    writer.flush()
}

class FastScanner(s: InputStream) {
    private var st = StringTokenizer("")
    private val br = BufferedReader(InputStreamReader(s))

    fun next(): String {
        while (!st.hasMoreTokens()) st = StringTokenizer(br.readLine())

        return st.nextToken()
    }

    fun nextInt() = next().toInt()
    fun nextLong() = next().toLong()
    fun nextLine() = br.readLine()
    fun nextDouble() = next().toDouble()
    fun ready() = br.ready()
}