from heapq import *

def process(arr,n):
    count = 0
    
    heap = []
    
    for i in range(len(arr)):
        heappush(heap,(arr[i],-(len(arr)-i))) # Constructing min-heap with second index as negative of maximum number of rides 
    
    while(n>0 and heap):
        cost,no_of_rides = heappop(heap)
        no_of_rides = -1 * no_of_rides # Changing maximum no_of_rides from negative to positive
        
        div = n//cost

        # If the amount of money is not sufficient to calculate the last number of rides user could take
        if(div<no_of_rides):
            count += div
            break

        # Else decrement the number of tokens by minimum cost * maximum no_of_rides
        else:
            count += no_of_rides
            n -= no_of_rides*cost
    

    return count;