p00060 Card Game
There is one card each with the numbers from "1" to "10", for a total of 10 cards. This card has numbers on the front and nothing on the back. Using this card, you and your opponent will play the game according to the following rules.

1. You and your opponent are dealt a total of two cards, one face up and one back up. You can see the numbers on the front card of your opponent, but not the numbers on the back card.
2. You win when the total number of cards dealt is 20 or less and greater than the total number of your opponent. For example, if your card is "7" "8" (15 total) and your opponent's card is "9" "10" (19 total), your opponent wins.
3. You and your opponent can draw up to one more card. You don't have to pull it.



Now, as a guide to deciding whether to draw one more card, consider the probability that the total will be 20 or less when you draw a card, and if that probability is 50% or more, draw a card. When calculating this probability, you can use the information of your two cards and the card on the opponent's table for a total of three cards. In other words, you don't draw those cards because you only have one for each card.

A program that reads your two cards and your opponent's front card, and outputs YES if there is a 50% or greater probability that the total will be 20 or less when you draw one more card, otherwise it will output NO. Please create.



Input

The input consists of multiple datasets. Given that the number on your first card is C1, the number on your second card is C2, and the number on your opponent's face card is C3, each dataset is given in the following format: ..


C1 C2 C3


Output

Print YES or NO on one line for each dataset.

Example

Input

1 2 3
5 6 9
8 9 10


Output

YES
YES
NO
```python
while 1:
    a=[i+1 for i in range(10)]
    try:b=list(map(int,input().split()))
    except:break
    for i in b:a.remove(i)
    print('YES' if sum([1 for i in a if i<=20-sum(b[:2])])/7>=0.5 else 'NO')
```