Problem:
Score : 300 points

Problem Statement

There are N people, conveniently numbered 1 through N.
They were standing in a row yesterday, but now they are unsure of the order in which they were standing.
However, each person remembered the following fact: the absolute difference of the number of the people who were standing to the left of that person, and the number of the people who were standing to the right of that person.
According to their reports, the difference above for person i is A_i.

Based on these reports, find the number of the possible orders in which they were standing.
Since it can be extremely large, print the answer modulo 10^9+7.
Note that the reports may be incorrect and thus there may be no consistent order.
In such a case, print 0.

Constraints

1≦N≦10^5

0≦A_i≦N-1

Input

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

N
A_1 A_2 ... A_N

Output

Print the number of the possible orders in which they were standing, modulo 10^9+7.

Sample Input 1

5
2 4 4 0 2

Sample Output 1

4

There are four possible orders, as follows:

2,1,4,5,3

2,5,4,1,3

3,1,4,5,2

3,5,4,1,2

Sample Input 2

7
6 4 0 2 4 0 2

Sample Output 2

0

Any order would be inconsistent with the reports, thus the answer is 0.

Sample Input 3

8
7 5 1 1 7 3 5 3

Sample Output 3

16

Solution in C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace ConsoleApplication14
{


    class Program
    {
        static void Main(string[] args)
        {
            int N = int.Parse(Console.ReadLine());
            string[] msg = Console.ReadLine().Split(' ');
            bool isOdd = ((N & 0x00000001) == 1) ? true : false;
            Dictionary<int, int> dic = new Dictionary<int, int>();

            //各絶対値は
            //人数が偶数ならば1～(N-1)のペア(ただし１つとび)
            //人数が奇数ならば0を１つ含む2～(N-1)のペア(ただし一つとび)
            //で構成されるはず
            //逸脱の場合は0人と判定してよい
            bool valid = true;
            for (int i = 0; i < N; i++)
            {
                int num = int.Parse(msg[i]);
                if (num >= N) { valid = false; break; }
                if ((isOdd) && ((num & 0x00000001) != 0)) { valid = false; break; }
                if ((!isOdd) && ((num & 0x00000001) == 0)) { valid = false; break; }

                if (dic.ContainsKey(num)) dic[num]++;
                else dic.Add(num, 1);

                if (dic[num] > 2) { valid = false; break; }
            }

            //一応辞書のカウントが合っているかどうかをザックリ見る
            if ((isOdd) && (dic.Count != ((N / 2) + 1))) { valid = false; }
            if ((!isOdd) && (dic.Count != (N / 2))) { valid = false; }

            //Nが奇数の場合に限り0という絶対値が一回だけ出現するはず
            if (isOdd){
                if (!dic.ContainsKey(0)) valid = false;
                else if (dic[0] != 1) valid = false;
            }

            if(!valid)
            {
                //該当なし
                Console.WriteLine("0");
            }
            else
            {
                //桁数より一意に組み合わせ数が確定するはず
                int val = (N / 2);
                UInt64 result = (UInt64)Math.Pow(2, val);
                result = (result % 1000000007);
                Console.WriteLine(result.ToString());
            }

            Console.ReadLine();
        }
    }
}