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: Accepted)

#include <bits/stdc++.h>
#define ll long long
#define MODV 1000000007
#define INF INT_MAX // 2147483647
#define INFLL LLONG_MAX // 9223372036854775807
#define rep(i, n) for(ll i=0, i##_len=(ll)(n); i<i##_len; i++)
#define repf(i, n) for(ll i=1, i##_len=(ll)(n+1); i<i##_len; i++)
#define per(i, n) for(ll i=((ll)(n))-1; i>=0; i--)
#define perf(i, n) for(ll i=((ll)(n)); i>0; i--)
#define all(v) v.begin(), v.end()
#define endl "\n"
#define vi vector<ll>
#define vvi vector<vector<ll>>
#define Yes() cout << "Yes" << endl
#define YES() cout << "YES" << endl
#define No() cout << "No" << endl
#define NO() cout << "NO" << endl
#define Init() std::ios::sync_with_stdio(false); std::cin.tie(0); std::cout<<fixed<<setprecision(15);
template<class T>bool chmax(T &a, const T &b){ if(a<b){ a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b){ if(b<a){ a=b; return 1; } return 0; }
using namespace std;

long long modPow(long long x, long long n, long long mod = MODV){
    if(n == 0) return 1;
    if(n%2 == 0) return modPow(x*x%mod, n/2, mod)%mod;
    return x%mod * modPow(x%mod, n-1, mod)%mod;
}

int main(){
    Init();
    ll n; cin >> n;
    // nが偶数だったら入力値は奇数、nが奇数だったら入力値は偶数でなければいけない
    ll chk = (n%2 == 0) ? 1 : 0;

    // 入力処理
    map<ll, ll> cnt;
    rep(i, n){
        ll in; cin >> in;
        cnt[in]++;
    }

    ll exp = 0;
    for(auto p: cnt){
        // 偶奇が一致しない場合はループから脱出
        if(p.first%2 != chk){ n = -1; break; }
        // 奇数長の場合で"0"の数が1以外の時はループから脱出、一致するなら無視
        else if(p.first == 0 && p.second != 1){ n = -1; break;}
        else if(p.first == 0 && p.second == 1){ continue; }
        // 2回以上同じ数字が出てきてもダメ
        else if(p.first != 0 && p.second != 2) { n = -1; break;}
        else exp++;
    }

    if(n == -1) cout << 0 << endl;
    else cout << modPow(2, exp, MODV) << endl;
}