Problem: 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.

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.

Examples

Input

1
10


Output

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


Input

2
1000


Output

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


Input

2
1001


Output

Impossible


Input

1
11


Output

Possible
0
1 1

Solution (in PYTHON3):

Solution:
import sys
from heapq import *

def first_1(x):
  if x:
    i = 0
    while x & 1 == 0:
      x >>= 1
      i += 1
    return i

def sep_single(s):
  if s:
    r = 1 << (s.bit_length()-1)
    i = s.bit_length()-1
    while s:
      if r & s:
        yield -i, r
        s ^= r
      r >>= 1
      i -= 1

def prepare_to_right(g, x):
  x0, y0 = g[-1]
  if x0 < x and y0:
    y0 = 0
    g.append((x0, 0))
  while x0 < x:
    x0 += 1
    g.append((x0, 0))

def prepare_to_left(g, x):
  x0, y0 = g[-1]
  if x+1 < x0 and y0:
    y0 = 0
    g.append((x0, 0))
  while x+1 < x0:
    x0 -= 1
    g.append((x0, 0))

def append_to_right(g, x, y):
  prepare_to_right(g, x)
  x0, y0 = g[-1]
  if y != y0:
    g.append((x, y))
  g.append((x+1, y))

def subloop_tp_group(g):
  x = 0
  while g&1 == 0:
    g >>= 1
    x += 1
  loop = [(x+1, 0)]
  tail = [(x, 0), (x, 1), (x+1, 1)]
  g >>= 1
  x += 1
  while g:
    while g&1==0:
      g >>= 1
      x += 1
    prepare_to_right(loop, x)
    prepare_to_right(tail, x)
    rev = reversed(loop)
    append_to_right(loop, x, 1)
    loop.append((x+1, 0))
    loop.extend(rev)
    loop.extend(tail)
    append_to_right(loop, x, 0)
    append_to_right(tail, x, 1)
    g >>= 1
    x += 1
  loop.extend(reversed(tail))
  return loop

N = int(input())

single = 0
tpg = []
for i in range(1<<N):
  a = int(sys.stdin.read(1))
  if bin(i).count("1") == 1:
    if a == 0:
      single |= i
  else:
    b = i & single
    if i and b | a == 0 and all(p & i != p for n, p in tpg):
      heappush(tpg, (-first_1(i), i))
    elif i | a == 0 or a and (b or any(p & i == p for n, p in tpg)):
      print("Impossible")
      break
else:
  print("Possible")
  loop = []
  for n, g in merge(sep_single(single), nsmallest(len(tpg),tpg)):
    subloop = subloop_tp_group(g)
    if loop:
      prepare_to_left(loop, subloop[0][0])
      if loop[-1] == subloop[0]:
        loop.pop()
    loop.extend(subloop)
  
  if not loop:
    loop.append((0, 0))
  prepare_to_right(loop, loop[0][0])
  
  print(len(loop)-1)
  for p in loop:
    print(*p)