CF768E.Game of Stones

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 s__i 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.

山姆一直在教琼恩玩“石子游戏”，以锻炼他的思维，并帮助他制定对抗异鬼的策略。该游戏规则非常简单：

• 游戏开始时有 $n$ 堆石子，编号从 $1$ 到 $n$。第 $i$ 堆包含 $s_i$ 颗石子。
• 双方轮流进行操作。一次操作定义为从某一堆石子中取走若干颗石子（取走 $0$ 颗石子不视为一次有效操作）。
• 无法进行操作的一方判负。

现在琼恩认为自己已准备好迎战，但山姆并不认同。为了证明自己的观点，山姆提议他们玩一个该游戏的变种版本。

在此变种版本中，对同一堆石子，不允许重复执行完全相同的操作。例如，若曾从某堆中取走 $4$ 颗石子，则此后不能再从该堆中取走 $4$ 颗石子。

山姆布置好游戏并率先行动。琼恩认为山姆只是在阻止自己奔赴战场。琼恩想知道：若双方均采取最优策略，自己是否能够获胜？

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

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

第一行包含一个整数 $n$（$1 \leq n \leq 10^6$）—— 表示石堆的数量。

接下来的 $n$ 行，每行包含一个整数 $s_i$（$1 \leq s_i \leq 60$）—— 表示第 $i$ 堆石子的数量。

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

如果琼获胜，则输出一行“YES”（不带引号），否则输出“NO”（不带引号）

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:

在第一种情况下，山姆取走所有石子，乔恩输掉游戏。

在第二种情况下，山姆可能的走法如下：

在上述每种情况下，乔恩均可通过最后一步获胜，具体如下：