Petr and a Combination Lock

题意翻译

有一个锁,它只有指针再次指到0刻度处才可以开锁(起始状态如图所示,一圈360度) 以下给出n个操作 及每次转动度数,如果可以通过逆时针或顺时针再次转到0度输出YES,否则输出NO

题目描述

Petr has just bought a new car. He's just arrived at the most known Petersburg's petrol station to refuel it when he suddenly discovered that the petrol tank is secured with a combination lock! The lock has a scale of $ 360 $ degrees and a pointer which initially points at zero: ![](https://cdn.luogu.com.cn/upload/vjudge_pic/CF1097B/390f775757c241d3e93f871107696e4a8b8d123d.png)Petr called his car dealer, who instructed him to rotate the lock's wheel exactly $ n $ times. The $ i $ -th rotation should be $ a_i $ degrees, either clockwise or counterclockwise, and after all $ n $ rotations the pointer should again point at zero. This confused Petr a little bit as he isn't sure which rotations should be done clockwise and which should be done counterclockwise. As there are many possible ways of rotating the lock, help him and find out whether there exists at least one, such that after all $ n $ rotations the pointer will point at zero again.

输入输出格式

输入格式


The first line contains one integer $ n $ ( $ 1 \leq n \leq 15 $ ) — the number of rotations. Each of the following $ n $ lines contains one integer $ a_i $ ( $ 1 \leq a_i \leq 180 $ ) — the angle of the $ i $ -th rotation in degrees.

输出格式


If it is possible to do all the rotations so that the pointer will point at zero after all of them are performed, print a single word "YES". Otherwise, print "NO". Petr will probably buy a new car in this case. You can print each letter in any case (upper or lower).

输入输出样例

输入样例 #1

3
10
20
30

输出样例 #1

YES

输入样例 #2

3
10
10
10

输出样例 #2

NO

输入样例 #3

3
120
120
120

输出样例 #3

YES

说明

In the first example, we can achieve our goal by applying the first and the second rotation clockwise, and performing the third rotation counterclockwise. In the second example, it's impossible to perform the rotations in order to make the pointer point at zero in the end. In the third example, Petr can do all three rotations clockwise. In this case, the whole wheel will be rotated by $ 360 $ degrees clockwise and the pointer will point at zero again.