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#testStr = '#ab!c#de!f'
#testStr = '##!r#po#'
testStr ='#a!b#c'
cnt=0
head = -1
tail = -1
ans = 0
for i in range(len(testStr)):
    if testStr[i]=='#':
        cnt+=1

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#!/bin/python3

import math
import os
import random
import re
import sys

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# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:     
    def findTarget(self, root: Optional[TreeNode], k: int) -> bool:
        '''
        print(root.val)

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class Solution:
    def complexNumberMultiply(self, num1: str, num2: str) -> str:
        #print(num1.split("+"))
        #print(complex(int(num1[0]),int(num1[0].split("i")[0])))
        
        a = complex(int(num1.split("+")[0]),int(num1.split("+")[1].split("i")[0]))
        b = complex(int(num2.split("+")[0]),int(num2.split("+")[1].split("i")[0]))
        print(a)
        print(b)
        #print(a*b)

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class Node:
  def __init__(self):
    self.data = None
    self.next = None

class LinkedList:
  def __init__(self):
    self.head = None

  def addNode(self, data):

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class Solution:
    def permute(self, nums: List[int]) -> List[List[int]]:
        ans = []
        # base
        if len(nums)==1:
            return [nums[:]] 
        
        for i in range(len(nums)): # counter
            tmp=nums.pop(0)
            res=self.permute(nums)

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class Solution:
    def isValidSerialization(self, preorder: str) -> bool:
        
        #print(preorder)
        
        # 一個element一個element加  
        ## 過程中 遇到 (number # #) => #
        ## 如果都沒有 (number # #)，則繼續一個 element 一個 element 加
        
        stack_list=preorder.split(",")

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class Solution:
    def permuteUnique(self, nums: List[int]) -> List[List[int]]:
        ans = []
        # base
        if len(nums)==1:
            return [nums[:]] 
        
        for i in range(len(nums)): # counter
            tmp=nums.pop(0)
            res=self.permuteUnique(nums)

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class Solution:
    def multiply(self, num1: str, num2: str) -> str:
        if num1=="0" or num2=="0":
            return "0"
        #print(num1[::-1])
        n1 = []
        n2 = []
        ans = []
        for i in num1[::-1]:
            n1.append(int(i))

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class Solution:
   def solve(self, times): 
      times.sort(key=lambda x: x[1])
      print(times)
      counter = 0
      end = -1

      for i in range(len(times)):
         print("start:",times[i][0])
         print("dy. end:",end)