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Time Limit: 2 sec / Memory Limit: 1024 MB

### 問題文

• x = \sum_{i \in S} 2^i とする。また、点集合 B_S\{ (i+0.5,0.5) | i \in S \} と定義する。
• 閉曲線 CB_S を通らずに連続的に動かして、閉曲線上のすべての点の y 座標を負にするような方法が存在するならば、A_x = 1 である。
• 閉曲線 CB_S を通らずに連続的に動かして、閉曲線上のすべての点の y 座標を負にするような方法が存在しないならば、A_x = 0 である。

### 補足

C が閉曲線であるとは、次のように定義される。

• C[0,1] から \mathbb{R}^2 への連続関数であり、C(0) = C(1) を満たす。

• 次の条件を満たす関数 f : [0,1] \times [0,1] \rightarrow \mathbb{R}^2 が存在する。
• f は連続。
• f(0,t) = C(t)
• f(1,t) = D(t)
• f(x,t) \notin X

### 制約

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

### 入力

N
A_0A_1 \cdots A_{2^N-1}


### 出力

L
x_0 y_0
x_1 y_1
:
x_L y_L


これは、閉曲線として (x_0,y_0),(x_1,y_1),\ldots,(x_L,y_L) をこの順に通る閉折れ線を選ぶことを意味する。

• 0 \leq x_i \leq N, 0 \leq y_i \leq 1, x_i,y_i は整数 \quad (0 \leq i \leq L)
• |x_i-x_{i+1}| + |y_i-y_{i+1}| = 1 \quad (0 \leq i \leq L-1)
• (x_0,y_0) = (x_L,y_L)

また、閉曲線の長さ L は、 0 \leq L \leq 250000 を満たす必要がある。

もとの問題で条件を満たす閉曲線が存在するならば、この出力形式のもとでも存在することが証明できる。

### 入力例 1

1
10


### 出力例 1

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


S = \emptyset のときは閉曲線上のすべての点の y 座標を負にすることが可能です。

S = \{0\} のときはどのようにしても閉曲線上のすべての点の y 座標を負にすることはできません。

### 入力例 2

2
1000


### 出力例 2

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


### 入力例 3

2
1001


### 出力例 3

Impossible


### 入力例 4

1
11


### 出力例 4

Possible
0
1 1


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