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 Ruby (2.3.3) (judge verdict: Wrong Answer)

# util functions

# debug
$debugFlag = ( ARGV[0] != nil )
p "==== DEBUG MODE ====" if $debugFlag

def pd(str)
  p str if $debugFlag
end

def printd(str)
  print str if $debugFlag
end

# input
inputs = []
lc = 0
while l = $stdin.gets do
  if false
    inputs << l.chomp.split(" ").map(&:to_s)
  else
    inputs << l.chomp.split(" ").map(&:to_i)
  end
  lc += 1
end

nn = inputs[0][0]
lst = inputs[1]

# input check
if $debugFlag
  p "--- INPUT (begin) ---"
  inputs.each do |l_i|
    p l_i
  end
  p "--- INPUT (end) ---"
  print "\n"
end

# body
if $debugFlag
  p nn
  p lst
end

flag = true
if nn % 3 != 0
  val = lst[0]
  lst.each do |l_i|
    if l_i != val
      flag = false
      break
    end
  end
  if nn % 2 != 0
    if val != 0
      flag = false
    end
  end
else
  cnt = 0
  vals_hh = {}
  lst.each do |l_i|
    if vals_hh[l_i] == nil
      if cnt == 3
        flag = false
        break
      else
        cnt += 1
        vals_hh[l_i] = 1
      end
    else
      vals_hh[l_i] += 1
    end
  end # l_i
  arr = vals_hh.to_a
  pd arr
  if arr.size == 1
    if !(arr[0][0] == 0)
      flag = false
    end
  elsif arr.size == 2
    if !((arr[0][0] == 0 or arr[1][0] == 0) and arr[0][1] == arr[1][1])
      flag = false
    end
  elsif !(arr[0][0] ^ arr[1][0] ^ arr[2][0] == 0 and arr[0][1] == arr[1][1] and arr[1][1] == arr[2][1])
    flag = false
  end
end

if flag == true
  puts "Yes"
else
  puts "No"
end