Problem:
Score : 300 points

Problem Statement

Snuke has N hats. The i-th hat has an integer a_i written on it.

There are N camels standing in a circle.
Snuke will put one of his hats on each of these camels.

If there exists a way to distribute the hats to the camels such that the following condition is satisfied for every camel, print Yes; otherwise, print No.

The bitwise XOR of the numbers written on the hats on both adjacent camels is equal to the number on the hat on itself.

What is XOR?

The bitwise XOR x_1 \oplus x_2 \oplus \ldots \oplus x_n of n non-negative integers x_1, x_2, \ldots, x_n is defined as follows:

- When x_1 \oplus x_2 \oplus \ldots \oplus x_n is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if the number of integers among x_1, x_2, \ldots, x_n whose binary representations have 1 in the 2^k's place is odd, and 0 if that count is even.

For example, 3 \oplus 5 = 6.

Constraints

All values in input are integers.

3 \leq N \leq 10^{5}

0 \leq a_i \leq 10^{9}

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 \ldots a_{N}

Output

Print the answer.

Sample Input 1

3
1 2 3

Sample Output 1

Yes

If we put the hats with 1, 2, and 3 in this order, clockwise, the condition will be satisfied for every camel, so the answer is Yes.

Sample Input 2

4
1 2 4 8

Sample Output 2

No

There is no such way to distribute the hats; the answer is No.

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

int N;
int a[200000];

#define MAGIC 4

int main(void) {
	int i, j;
	int kazu[MAGIC] = {0};
	int count[MAGIC] = {0};
	int slot = 0;
	int yes;
	if (scanf("%d", &N) != 1) return 1;
	for (i = 0; i < N; i++) {
		if (scanf("%d", &a[i]) != 1) return 1;
	}
	for (i = 0; i < N; i++) {
		int found = 0;
		for (j = 0; j < MAGIC && j < slot; j++) {
			if (kazu[j] == a[i]) {
				found = 1;
				count[j]++;
			}
		}
		if (!found && slot < MAGIC) {
			kazu[slot] = a[i];
			count[slot] = 1;
			slot++;
		}
	}
	yes = 0;
	switch (slot) {
		case 1:
			/* 1種類→全部0ならおｋ、それ以外はNG */
			yes = 0;
			if (kazu[0] == 0) yes = 1;
			break;
		case 2:
			/* 2種類→3種類のx 0 x相当ならおｋ，それ以外はNG */
			yes = 0;
			if (kazu[0] == 0 && count[0] * 2 == count[1]) yes = 1;
			if (kazu[1] == 0 && count[1] * 2 == count[0]) yes = 1;
			break;
		case 3:
			yes = 1;
			if (count[0] != count[1] || count[1] != count[2]) yes = 0; /* 数が同じでないとダメ */
			if ((kazu[0] ^ kazu[1]) != kazu[2]) yes = 0; /* xor関係になってないとダメ */
			break;
		default:
			yes = 0;
	}
	puts(yes ? "Yes" : "No");
	return 0;
}

/*

x1 + x3 = x2
x2 + x1 = x3
x3 + x2 = x1

x1 + x3 = x2
x2 + x4 = x3
...
xi-2 + xi = xi-1
xi-1 + xi+1 = xi
xi + xi+2 = xi+1
...
xj-2 + xj = xj-1
xj-1 + xj+1 = xj
xj + xj+2 = xj+1
...
xn-1 + x1 = xn
xn + x2 = x1

×全部xorして0ならYes
4
2 2 2 2

------
1 2 3 (1 2)
1 3 2 (1 3)
2 1 3 (2 1)
2 3 1 (2 3)
3 1 2 (3 1)
3 2 1 (3 2)

------

00010
00101
00111
00010
00101
00111

最初に決めた2個+そこから決まる1個でループになる

スタート
↓
Nが3の倍数でない→No
分布が3通りの数を同じ数ずつでない→No
3通りの数がxor関係にない→No
↓
Yes

a + b = c のとき、両辺にbをxorして a = c + b
両辺にaをxorして b = c + a
よって、xor関係は1個チェックすればおｋ

*/