Little Vlad is fond of popular computer game Bota-2. Recently, the developers announced the new add-on named Bota-3. Of course, Vlad immediately bought only to find out his computer is too old for the new game and needs to be updated.

There are *n* video cards in the shop, the power of the *i*-th video card is equal to integer value *a*<sub class="lower-index">*i*</sub>. As Vlad wants to be sure the new game will work he wants to buy not one, but several video cards and unite their powers using the cutting-edge technology. To use this technology one of the cards is chosen as the leading one and other video cards are attached to it as secondary. For this new technology to work it's required that the power of each of the secondary video cards is divisible by the power of the leading video card. In order to achieve that the power of any secondary video card can be reduced to any integer value less or equal than the current power. However, the power of the leading video card should remain unchanged, i.e. it can't be reduced.

Vlad has an infinite amount of money so he can buy any set of video cards. Help him determine which video cards he should buy such that after picking the leading video card and may be reducing some powers of others to make them work together he will get the maximum total value of video power.
The first line of the input contains a single integer *n* (1 ≤ *n* ≤ 200 000) — the number of video cards in the shop.

The second line contains *n* integers *a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub> (1 ≤ *a*<sub class="lower-index">*i*</sub> ≤ 200 000) — powers of video cards.
In the first sample, it would be optimal to buy video cards with powers 3, 15 and 9. The video card with power 3 should be chosen as the leading one and all other video cards will be compatible with it. Thus, the total power would be 3 + 15 + 9 = 27. If he buys all the video cards and pick the one with the power 2 as the leading, the powers of all other video cards should be reduced by 1, thus the total power would be 2 + 2 + 14 + 8 = 26, that is less than 27. Please note, that it's not allowed to reduce the power of the leading video card, i.e. one can't get the total power 3 + 1 + 15 + 9 = 28.

In the second sample, the optimal answer is to buy all video cards and pick the one with the power 2 as the leading. The video card with the power 7 needs it power to be reduced down to 6. The total power would be 8 + 2 + 2 + 6 = 18.
4
3 2 15 9
27
4
8 2 2 7
18
1
1
1
1
123819
123819
10
9 6 8 5 5 2 8 9 2 2
52
100
17 23 71 25 50 71 85 46 78 72 89 26 23 70 40 59 23 43 86 81 70 89 92 98 85 88 16 10 26 91 61 58 23 13 75 39 48 15 73 79 59 29 48 32 45 44 25 37 58 54 45 67 27 77 20 64 95 41 80 53 69 24 38 97 59 94 50 88 92 47 95 31 66 48 48 56 37 76 42 74 55 34 43 79 65 82 70 52 48 56 36 17 14 65 77 81 88 18 33 40
5030
100
881 479 355 759 257 497 690 598 275 446 439 787 257 326 584 713 322 5 253 781 434 307 164 154 241 381 38 942 680 906 240 11 431 478 628 959 346 74 493 964 455 746 950 41 585 549 892 687 264 41 487 676 63 453 861 980 477 901 80 907 285 506 619 748 773 743 56 925 651 685 845 313 419 504 770 324 2 559 405 851 919 128 318 698 820 409 547 43 777 496 925 918 162 725 481 83 220 203 609 617
50692
12
2 3 5 5 5 5 5 5 5 5 5 5
50
a=int(input())
s1=[]
s2=[]
s3=[]
s1=list(map(int,input().split()))
li=sorted(s1)
for x in range(a):
    max1=0
    if x in s3:
        continue
    s3.append(x)
    for y in range(a):
        if li[y]%li[x]!=0:
            yu=li[y]%li[x]
            max1=max1+li[y]-yu
        else:
            max1=max1+li[y]
    s2.append(max1)
print(max(s2))
 	 		 	  			 		 					 	    	 		