imedadel

def pickingNumbers(arr):
    maximum = 0
    for number in arr:
        freq = arr.count(number) + arr.count(number-1)
        if freq > maximum:
            maximum = freq
    return maximum

imedadel

def getTotalX(a, b):
    maxN = max(b)
    minN = min(a)
    sumN = 0
    for number in range(minN, maxN+1):
        factorsOfNumber = all([number%x == 0 for x in a])
        multsOfNumber = all([x%number == 0 for x in b])
        sumN += factorsOfNumber * multsOfNumber
    return sumN

imedadel

toys, budget = [int(x) for x in input().strip().split()]

spent = 0
bought = 0

prices = [int(x) for x in input().strip().split()]
prices.sort()

for price in prices:
    if spent + price <= budget:

imedadel

q = int(input())

while(q>0):
    q -= 1
    letters = [c for c in input().strip()]
    cleaned = []
    cleaned.append(letters[0])
    for letter in letters[1:]:
        if letter != cleaned[-1]:
            cleaned.append(letter)

imedadel

def countSwaps(a):
    swaps = 0
    n = len(a)
    for i in range(n):
        for j in range(n-1):
        # Swap adjacent elements if they are in decreasing order
            if a[j] > a[j + 1]:
                a[j], a[j + 1] = a[j+1], a[j]
                swaps += 1
    print("Array is sorted in", swaps, "swaps.")

imedadel

#!/bin/python3

import math
import os
import random
import re
import sys

first_multiple_input = input().rstrip().split()

imedadel

from collections import Counter
def freqQuery(queries):
    items = Counter()
    checks = Counter()
    res = []
    for couple in queries:
        query, item = couple
        if query == 1:
            checks[items[item]] -= 1
            items[item] += 1

imedadel

def binary_search(list, item):
  low = 0
  high = len(list)-1
  
  while low <= high:
    mid = (low + high) // 2
    guess = list[mid]
    if guess == item:
      return mid
    elif guess > item:

imedadel

pairs = int(input())

from collections import Counter

def checkPairs(first, second):
    firstCount = Counter(first)
    secondCount = Counter(second)
    commonLettersExist = firstCount - secondCount != firstCount
    if commonLettersExist:
        return "YES"

imedadel

# Using Counter
# see https://www.journaldev.com/20806/python-counter-python-collections-counter#python-counter-arithmetic-operations
# (Counter(note) - Counter(magazine)) returns only the elements with a positive count
# That is, if there are more less in `note` than `magazine`, we would get an empty dictionary.
from collections import Counter
def checkMagazine(magazine, note):
    if (Counter(note) - Counter(magazine)) == {}:
        print("Yes")
    else:
        print("No")