Problem:
Score : 300 points

Problem Statement

There are N integers written on a blackboard. The i-th integer is A_i.

Takahashi will repeatedly perform the following operation on these numbers:

Select a pair of integers, A_i and A_j, that have the same parity (that is, both are even or both are odd) and erase them.

Then, write a new integer on the blackboard that is equal to the sum of those integers, A_i+A_j.

Determine whether it is possible to have only one integer on the blackboard.

Constraints

2 ≦ N ≦ 10^5

1 ≦ A_i ≦ 10^9

A_i is an integer.

Input

The input is given from Standard Input in the following format:

N
A_1 A_2 … A_N

Output

If it is possible to have only one integer on the blackboard, print YES. Otherwise, print NO.

Sample Input 1

3
1 2 3

Sample Output 1

YES

It is possible to have only one integer on the blackboard, as follows:

Erase 1 and 3 from the blackboard, then write 4. Now, there are two integers on the blackboard: 2 and 4.

Erase 2 and 4 from the blackboard, then write 6. Now, there is only one integer on the blackboard: 6.

Sample Input 2

5
1 2 3 4 5

Sample Output 2

NO

Solution in Ruby (2.3.3) (judge verdict: Wrong Answer)

k = gets.to_i
a = ""
a = gets.chomp.split(" ").map(&:to_i)

count_kisu = 0
count_gusu = 0
len = 0

if(k%2 == 0)
   len = (k / 2) - 1
else    
    len = k / 2
end

for i in 0..len
    if(a[i]%2 != 0)
        count_kisu += 1
    else
        count_gusu += 1
    end
        
    if(i != k-1-i)
        if(a[k-1-i]%2 != 0)
            count_kisu += 1
        else
            count_gusu += 1
        end
    end
end

while count_gusu%2 == 0
    count_gusu = count_gusu / 2
    if(count_gusu == 0)
        flg = 1
        break
    end
end

if(count_kisu%2 != 0)
    print("NO")
elsif((count_kisu%2 == 0) && (k%2 == 0) && (flg == 1))
    print("NO")
else
    print("YES")
end