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 Rust (1.15.1) (judge verdict: Wrong Answer)

#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::*;
use std::io::{Write, BufWriter};
// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
macro_rules! input {
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, [ $t:tt ]) => {{
        let len = read_value!($next, usize);
        (0..len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    }};

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

#[allow(unused)]
macro_rules! debug {
    ($($format:tt)*) => (write!(std::io::stderr(), $($format)*).unwrap());
}
#[allow(unused)]
macro_rules! debugln {
    ($($format:tt)*) => (writeln!(std::io::stderr(), $($format)*).unwrap());
}

fn solve() {
    let out = std::io::stdout();
    let mut out = BufWriter::new(out.lock());
    macro_rules! puts {
        ($($format:tt)*) => (let _ = write!(out,$($format)*););
    }
    input! {
        n: usize,
        a: chars,
    }
    for i in 0..1 << n {
        for j in 0..1 << n {
            if (i & j) == i && a[i] == '0' && a[j] == '1' {
                puts!("Impossible\n");
                return;
            }
        }
    }
    if a[0] == '0' {
        puts!("Impossible\n");
        return;
    }
    let mut ans = vec![];
    for i in 1..1 << n {
        if a[i] == '1' {
            continue;
        }
        if i.is_power_of_two() {
            // one loop
            let x = (i - 1).count_ones() as i64;
            for j in 0..x + 1 {
                ans.push((j, 0));
            }
            ans.push((x, 1));
            ans.push((x + 1, 1));
            for j in (1..x + 2).rev() {
                ans.push((j, 0));
            }
            continue;
        }
        let mut c = vec![];
        for j in 0..n {
            if (i & 1 << j) != 0 {
                c.push(j);
            }
        }
        let &ma = c.last().unwrap();
        for j in 0..ma + 1 {
            if j != ma && (i & 1 << j) != 0 {
                let j = j as i64;
                ans.push((j, 0));
                ans.push((j, 1));
                ans.push((j + 1, 1));
            } else {
                ans.push((j as i64, 0));
            }
        }
        for j in (0..ma + 1).rev() {
            let x = j as i64;
            if (i & 1 << j) != 0 {
                ans.push((x + 1, 0));
                ans.push((x + 1, 1));
                ans.push((x, 1));
            } else {
                ans.push((x + 1, 0));
            }
        }
        for j in 0..ma {
            ans.push((j as i64, 0));
        }
        let max = ma as i64;
        ans.push((max, 0));
        ans.push((max, 1));
        ans.push((max + 1, 1));
        for j in (0..ma + 1).rev() {
            ans.push((j as i64 + 1, 0));
        }
    }
    ans.push((0, 0));
    assert!(ans.len() <= 250_000);
    puts!("Possible\n");
    puts!("{}\n", ans.len());
    for (x, y) in ans {
        puts!("{} {}\n", x, y);
    }
}

fn main() {
    // In order to avoid potential stack overflow, spawn a new thread.
    let stack_size = 104_857_600; // 100 MB
    let thd = std::thread::Builder::new().stack_size(stack_size);
    thd.spawn(|| solve()).unwrap().join().unwrap();
}