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# https://docs.python.org/3.4/library/turtle.html
from turtle import *
import random

def draw_cookie(x, y):
    penup()
    goto(x,y)
    pendown()
    radius = random.randint(100,200)
    col = random.choice(['#CC6600','#FFCD82','#8F4700'])

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def iterate_by_chunk(word, chunk_length):
    i = 0
    while i < len(word):
        yield word[i:i+chunk_length]
        i += chunk_length


def check(word, pattern, len_pattern_word, i_offset):
    """check if the word match the pattern"""
    pattern_length = len(pattern)*len_pattern_word

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given n, print all the combinations of 0...(n-1) of length n

example

n = 2

00
01
10

11

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#https://code.google.com/codejam/contest/7214486/dashboard
from collections import deque
from itertools import combinations

class Matrix:
    @staticmethod
    def from_string(M):
        return Matrix([list(x.strip()) for x in M.split('\n')])
    
    def __init__(self,M):

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from pprint import pprint as pp
from collections import deque

def find_bridge_bf(nodes, edges, a, b):
    """
    Simple brute-force bridge finding
    For each edge, explore nodes from *a* and try to reach b without using the edge
    If it can't reach the *b*, it was a bridge
    """
    for edge in edges:

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#!/usr/bin/env python3
import sys

"""
Use a naive implementation of a Trie: http://en.wikipedia.org/wiki/Trie
"""
class Trie:

    def __init__(self):
        self.children = {} #letter<->sub-Trie mappings

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#Execute a type 3 grammar

rules2 = """
S -> ATC
A -> aA | 
C -> cC | 
T -> abT | ab
"""

class bcolors:

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import numpy as np
import sys
import random

MAX_ERASE_RATIO = 30  # 10 => 27 500
DILLATE_SIZE = 0

ERASED = -8
EXTRA = -6

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1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 1 1 0 1 1 0 0 1 1 0 0 1 0 0 1 0 1 0 0 0 1 0 1 1 1 1 0 0 1 1 1 0 1 0 0 0 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1
0 1 0 1 1 -1 1 0 -1 1 0 1 1 0 1 0 1 0 1 1 0 -1 1 0 1 -1 1 0 1 -1 1 1 1 -1 0 1 0 1 1 0 0 0 1 -1 0 -1 0 -1 1 0 0 -1 1
1 1 1 1 0 1 -1 1 1 1 0 1 -1 0 1 0 1 1 -1 0 1 1 -1 -1 1 0 -1 1 0 0 1 0 1 1 0 1 1 1 -1 -1 1 1 1 1 1 1 1 0 1 1 1 1 1
1 1 0 1 1 -1 0 1 0 1 0 1 -1 1 -1 1 0 1 1 1 1 1 0 0 -1 0 -1 1 1 1 -1 0 1 1 1 0 -1 1 1 0 1 0 1 1 1 1 1 1 0 0 -1 -1 1
-1 1 -1 -1 -1 1 1 -1 0 1 0 -1 1 1 -1 1 0 1 -1 1 0 -1 0 1 -1 0 0 -1 1 0 -1 0 -1 1 1 -1 0 -1 -1 0 0 -1 1 0 0 -1 -1 -1 -1 -1 0 1 1
0 1 0 0 1 0 0 1 1 -1 1 1 1 0 1 1 0 -1 0 1 1 1 1 1 0 0 1 1 1 1 0 1 -1 1 1 1 1 1 1 1 -1 1 1 1 1 1 1 1 1 1 0 0 -1
-1 0 -1 1 1 1 1 1 1 1 0 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 -1 1 1 -1 0 1 1 0 0 1 1 1 1 -1 -1 0 1 1 0 1 1 1 1 0 1 1
0 -1 0 -1 1 1 0 0 0 1 1 1 1 -1 1 1 1 1 1 1 1 1 1 0 -1 1 1 1 1 1 0 1 1 1 -1 1 1 0 1 1 1 0 1 1 1 1 1 0 -1 1 0 1 0
1 1 

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from __future__ import print_function
from itertools import *

def find_switches(a,b):
    return {i for i,v in enumerate(a) if v != b[i]}

def transform(a,switchs):
    r = ""
    for i, v in enumerate(a):
        if i in switchs: