#AGC043E. [AGC043E] Topology

[AGC043E] Topology

Score : 14001400 points

Problem Statement

Given are a positive integer NN and a sequence of length 2N2^N consisting of 00s and 11s: A0,A1,,A2N1A_0,A_1,\ldots,A_{2^N-1}. Determine whether there exists a closed curve CC that satisfies the condition below for all 2N2^N sets S{0,1,,N1}S \subseteq \{0,1,\ldots,N-1 \}. If the answer is yes, construct one such closed curve.

  • Let x=iS2ix = \sum_{i \in S} 2^i and BSB_S be the set of points {(i+0.5,0.5)iS}\{ (i+0.5,0.5) | i \in S \}.
  • If there is a way to continuously move the closed curve CC without touching BSB_S so that every point on the closed curve has a negative yy-coordinate, Ax=1A_x = 1.
  • If there is no such way, Ax=0A_x = 0.

For instruction on printing a closed curve, see Output below.

Notes

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

  • CC is a continuous function from [0,1][0,1] to R2\mathbb{R}^2 such that C(0)=C(1)C(0) = C(1).

We say that a closed curve CC can be continuously moved without touching a set of points XX so that it becomes a closed curve DD if and only if:

  • There exists a function f:[0,1]×[0,1]R2f : [0,1] \times [0,1] \rightarrow \mathbb{R}^2 that satisfies all of the following.- ff is continuous.
    • f(0,t)=C(t)f(0,t) = C(t).
    • f(1,t)=D(t)f(1,t) = D(t).
    • f(x,t)Xf(x,t) \notin X.
  • ff is continuous.
  • f(0,t)=C(t)f(0,t) = C(t).
  • f(1,t)=D(t)f(1,t) = D(t).
  • f(x,t)Xf(x,t) \notin X.

Constraints

  • 1N81 \leq N \leq 8
  • Ai=0,1(0i2N1)A_i = 0,1 \quad (0 \leq i \leq 2^N-1)
  • A0=1A_0 = 1

Input

Input is given from Standard Input in the following format:

NN

A0A1A2N1A_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:

LL

x0x_0 y0y_0

x1x_1 y1y_1

::

xLx_L yLy_L

This represents the closed polyline that passes (x0,y0),(x1,y1),,(xL,yL)(x_0,y_0),(x_1,y_1),\ldots,(x_L,y_L) in this order.

Here, all of the following must be satisfied:

  • 0xiN0 \leq x_i \leq N, 0yi10 \leq y_i \leq 1, and xi,yix_i, y_i are integers. (0iL0 \leq i \leq L)
  • xixi+1+yiyi+1=1|x_i-x_{i+1}| + |y_i-y_{i+1}| = 1. (0iL10 \leq i \leq L-1)
  • (x0,y0)=(xL,yL)(x_0,y_0) = (x_L,y_L).

Additionally, the length of the closed curve LL must satisfy 0L2500000 \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.

1
10
Possible
4
0 0
0 1
1 1
1 0
0 0

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

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

2
1000
Possible
6
1 0
2 0
2 1
1 1
0 1
0 0
1 0

The output represents the following curve:

2
1001
Impossible
1
11
Possible
0
1 1