Fox Ciel is playing a game with numbers now. 

Ciel has *n* positive integers: *x*<sub class="lower-index">1</sub>, *x*<sub class="lower-index">2</sub>, ..., *x*<sub class="lower-index">*n*</sub>. She can do the following operation as many times as needed: select two different indexes *i* and *j* such that *x*<sub class="lower-index">*i*</sub> &gt; *x*<sub class="lower-index">*j*</sub> hold, and then apply assignment *x*<sub class="lower-index">*i*</sub> = *x*<sub class="lower-index">*i*</sub> - *x*<sub class="lower-index">*j*</sub>. The goal is to make the sum of all numbers as small as possible.

Please help Ciel to find this minimal sum.
The first line contains an integer *n* (2 ≤ *n* ≤ 100). Then the second line contains *n* integers: *x*<sub class="lower-index">1</sub>, *x*<sub class="lower-index">2</sub>, ..., *x*<sub class="lower-index">*n*</sub> (1 ≤ *x*<sub class="lower-index">*i*</sub> ≤ 100).
import heapq

n = int(input())
numbers = [(-int(x), int(x)) for x in input().split()]
heapq.heapify(numbers)

while True:
    replaced_element = False
    readd_elements = []
    while len(numbers) >= 2 and not replaced_element:
        _, first_number = heapq.heappop(numbers)
        _, second_number = heapq.heappop(numbers)

        # print(first_number, second_number)

        if first_number > second_number:
            replaced_element = True
            first_number -= second_number
        
        readd_elements.append(first_number)
        readd_elements.append(second_number)
    
    for element in readd_elements:
        heapq.heappush(numbers, (-element, element))
    
    if not replaced_element:
        break

print(sum(number[1] for number in numbers))

 		  		     			 		  	 		   		 	