n = len(A)
    iterations=0
    i=0
    heapq.heapify(A)
    while i < n:
        minA = heapq.heappop(A) 
        if minA < k:
            if len(A)<1:
                return -1
            minB = heapq.heappop(A) 
            newValue= (1 * minA) + (2 * minB)
            heapq.heappush(A,newValue)
            i+=1
            iterations+=1
        else:
            break
    return iterations