import heapq

def min_cost_to_reduce_array(nums):
    heapq.heapify(nums)  # Convert the input list into a min-heap
    total_cost = 0

    while len(nums) > 1:
        # Extract the two minimum elements from the heap
        first = heapq.heappop(nums)
        second = heapq.heappop(nums)

        # Calculate the cost of the current move
        cost = first + second
        total_cost += cost

        # Add the sum back to the heap
        heapq.heappush(nums, cost)

    return total_cost