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)

#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll pow_quick(ll a, ll n)
{
    if(n == 0) return 1;
    ll x = pow_quick(a, n/2);
    long long ans = (long long)x * x;
    if(n % 2 == 1) ans = ans *a ;
    return ans;//
}
int sum[600000];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1; i<=n; i++)
    {
        int v;
        scanf("%d",&v);
        sum[v]++;
        if(sum[v]>=3)
        {
            printf("0\n");//不符合情况一
            return 0;
        }
    }
    if(n&1)
    {
        //非0
        if(sum[0]!=1)
        {
            //0只能出现一次
            printf("0\n");
            return 0;
        }
    }
    if(!(n%2))
    {
        //n为偶数
        if(sum[0])
        {
            //0的个数不为0不符合条件
            printf("0\n");
            return 0;
        }
    }
    printf("%lld\n",pow_quick(2,n/2));
    return 0;
}