Codeforces Round #399：E. Game of Stones

E. Game of Stones

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Sam has been teaching Jon the Game of Stones to sharpen his mind and help him devise a strategy to fight the white walkers. The rules of this game are quite simple:

• The game starts with n piles of stones indexed from 1 to n. The i-th pile contains si stones.
• The players make their moves alternatively. A move is considered as removal of some number of stones from a pile. Removal of 0 stones does not count as a move.
• The player who is unable to make a move loses.

Now Jon believes that he is ready for battle, but Sam does not think so. To prove his argument, Sam suggested that they play a modified version of the game.

In this modified version, no move can be made more than once on a pile. For example, if 4 stones are removed from a pile, 4 stones cannot be removed from that pile again.

Sam sets up the game and makes the first move. Jon believes that Sam is just trying to prevent him from going to battle. Jon wants to know if he can win if both play optimally.

Input

First line consists of a single integer n (1 ≤ n ≤ 106) — the number of piles.

Each of next n lines contains an integer si (1 ≤ si ≤ 60) — the number of stones in i-th pile.

Output

Print a single line containing "YES" (without quotes) if Jon wins, otherwise print "NO" (without quotes)

Examples

input

15

output

NO

input

212

output

YES

Note

In the first case, Sam removes all the stones and Jon loses.

In second case, the following moves are possible by Sam: 

In each of these cases, last move can be made by Jon to win the game as follows: 

翻译：有n堆石子，每次可以在其中一堆中取走任意数量的石子，但不能不取，并且对于同一堆，如果之前某一次取

走了x个，那么就再不能取走x个了，当一个人不能取了（当然石子可能还有剩余）的时候失败且游戏结束，问后者

是否能赢（是-YES）

仍然枚举前几个sg[x]，找规律

sg[0] = 0、sg[1] = 1;

sg[2] = mex{sg[0], sg[1]'} = mex{0,0} = 1;      ------      这里的sg[1]'表示已经取了一个石子还剩一个石子的状态，很

显然不能取！sg[1]'==0

sg[3] = mex{sg[0], sg[1]', sg[2]'} = mex{0,1,1} = 2;      ------      同上，很好证明sg[2]'==1，sg[1]'==1，不过注意这

里的sg[1]'表示已经取了两个石子还剩一个石子的状态，很显然是可以取的，这和上面的sg[1]'不一样，这里sg[1]'==1

sg[4] = mex{sg[0], sg[1]', sg[2]', sg[3]'} = mex{0,1,1,mex{0,sg[1]'}} = 2      ------      很显然这里的sg[1]'表示已经取过

一个石子也取过两个石子还剩一个石子的状态，很显然不能取，sg[1]'==0

……

最后可得出sg[x] = {0,1,1,2,2,2,3,3,3,3,4,4,4,4,4,5,5,5,5,5,5……}中秩为x的元素

#include<stdio.h>#include<algorithm>#include<vector>using namespace std;int n, ans;int main(void){int i, j, p;scanf("%d", &n);for(i=1;i<=n;i++){scanf("%d", &p);for(j=1;j*(j+1)/2<=p;j++);ans ^= j-1;}if(ans)printf("NO\n");elseprintf("YES\n");return 0;}