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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdbool.h>
#include<math.h>
long long f[100002];

int main() {
	int N;
	scanf("%d", &N);
	int i, A, tmp = 0, ans = 0;
	bool even = true;
	for (i = 0; i < N; i++)
	{
		scanf("%d", &A);
		f[A]++;
		//成立しているかチェックする
		if (i == 0 && A % 2 == 1)//f[0]が奇数の場合、全てが奇数か？
		{
			even = true;
		}
		else if(i == 0)
		{
			even = false;
		}
		if (even == true) {
			if (A % 2 != 1)//全てが奇数ではなかった場合
			{
				//ありうる並び方がないので０を出力する
				puts("0");
				return 0;
			}
			else//今のところ全てが奇数の場合
			{
				if (f[A] == 2)//２つあったものを数える
				{
					tmp++;
				}
				if (i == N - 1 && N / 2 == tmp)//最終的に全て２つあった場合
				{
					ans = (int)pow(2, ((double)N / 2));
					printf("%d\n", ans);
					return 0;
				}
			}
		}
		if (even != true) {
			if (A % 2 == 1 && A != 0)//ゼロで無い全てのものが偶数ではなかった場合
			{
				//ありうる並び方がないので０を出力する
				puts("0");
				return 0;
			}
			else if (f[0] > 1) //ゼロの発生回数が１つ以下ではなかった場合
			{
				//ありうる並び方ではないので０を出力する
				puts("0");
				return 0;
			}
			else//ゼロの発生回数が１つ以下で今のところゼロで無い全てのもの全てが偶数の場合
			{
				if (f[A] == 2)//２つあったものを数える
				{
					tmp++;
				}
				if (f[0] != 1 && i == N - 1)//すべてが数え終わったときゼロであったものは一つではなかった場合
				{
					//ありうる並び方ではないので０を出力する
					puts("0");
					return 0;
				}
				else if (i == N - 1 && (N - 1) / 2 == tmp && (N - 1) % 2 == 0 && f[0] == 1)//最終的にゼロで無い全てのもの全てが偶数かつゼロであったものは一つだった場合
				{
					ans = (int)pow(2, ((double)(N - 1) / 2));
					printf("%d\n", ans);
					return 0;
				}
			}
		}
	}
	return 0;
}