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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#pragma region _head

#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<cmath>
#include<map>
#include<set>
#include<iomanip>
#include<queue>
#include<stack>
#include<numeric>
#include<utility>
#include<regex>

#pragma region _define

#define f(i,a,b) for(int i=a;i<b;i++)
#define f_vI(v,n) f(i,0,n)cin>>v[i]
#define f_v2I(v1,v2,n) f(i,0,n)cin>>v1[i]>>v2[i]
#define f_v3I(v1,v2,v3,n) f(i,0,n)cin>>v1[i]>>v2[i]>>v3[i]
#define f_vO(v,n) f(i,0,n)cout<<v[i]<<endl
#define all(a) a.begin(),a.end()
#define size(s) s.size()
#define check() cout<<"! ! !"
#define endl "\n"
#define _y() cout<<"Yes"<<endl
#define _Y() cout<<"YES"<<endl
#define _n() cout<<"No"<<endl
#define _N() cout<<"NO"<<endl

#pragma endregion

#pragma region _using

using namespace std;

using LL = long long;
using st = string;
using vi = vector<LL>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using vc = vector<char>;
using vs = vector<st>;
using vb = vector<bool>;
using vvb = vector<vb>;
using vvvb = vector<vvb>;
using si = stack<LL>;
using ss = stack<st>;
using pi = pair<LL, LL>;
using v_pi = vector<pi>;

#pragma endregion

#pragma region _graph

#define node 50  //グラフの最大位数　適宜変更
//bool graph[node][node]; bool visited[node] = { false };
#define link(a,b,m) f(i,0,m){cin>>a[i]>>b[i]; a[i]--; b[i]--; graph[a[i]][b[i]]=graph[b[i]][a[i]]=true;}

#pragma endregion

LL gcd(LL a, LL b) {
	LL r;
	while ((r = a % b) != 0) {
		a = b;
		b = r;
	}
	return b;
}
LL lcm(LL a, LL b) {
	return (a / gcd(a, b) * b);
}

#pragma endregion 

/*****************************************************************************/
int main(void) {
	cin.tie(0); ios::sync_with_stdio(false);

	int n;
	cin >> n;
	vi a(n);
	f_vI(a, n);
	map<LL, LL>m;
	f(i, 0, n) {
		m[a[i]]++;
	}
	bool f = true;
	int i;
	if (n % 2 != 0) {
		i = 0;
	}
	else {
		i = 1;
	}
	while (i < n) {
		if (i == 0) {
			if (m[i] != 1) {
				f = false;
			}
		}
		else if (m[i] != 2) {
			f = false;
		}
		i += 2;
	}
	if (!f) {
		cout << 0;
	}
	else {
		cout << (LL)pow((LL)2, (LL)n / 2) % 1000000009;
	}
	return 0;
}