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 PI 3.14159265358979323846
#define MAXINF (1e18L)
#define INF (1e9L)
#define EPS (1e-9)
#define MOD ((ll)(1e9+7))
#define REP(i, n) for(int i=0;i<int(n);++i)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define RREP(i, n) for(int i=int(n)-1;i>=0;--i)
#define ALL(v) v.begin(),v.end()
#define FIND(v,x) (binary_search(ALL(v),(x)))
#define SORT(v) sort(ALL(v))
#define RSORT(v) sort(ALL(v));reverse(ALL(v))
#define DEBUG(x) cerr<<#x<<": "<<x<<endl;
#define DEBUG_VEC(v) cerr<<#v<<":";for(int i=0;i<v.size();i++) cerr<<" "<<v[i]; cerr<<endl
#define Yes(n) cout<<((n)?"Yes":"No")<<endl
#define YES(n) cout<<((n)?"YES":"NO")<<endl
#define pb push_back
#define fi first
#define se second
using namespace std;
template<class A>void pr(A a){cout << (a) << endl;}
template<class A,class B>void pr(A a,B b){cout << a << " "  ;pr(b);}
template<class A,class B,class C>void pr(A a,B b,C c){cout << a << " " ;pr(b,c);}
template<class A,class B,class C,class D>void pr(A a,B b,C c,D d){cout << a << " " ;pr(b,c,d);}
template<class T> inline bool chmin(T& a, T b){return a>b ? a=b, true : false;}
template<class T> inline bool chmax(T& a, T b){return a<b ? a=b, true : false;}
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll,ll> pll;

template<uint32_t mod>
struct ModInt{
    uint32_t a;
    ModInt& s(uint32_t vv){
        a=vv<mod?vv:vv-mod;
        return *this;
    }
 
    ModInt(int64_t x=0){s(x%mod+mod);}
 
    ModInt& operator+=(const ModInt &x){return s(a+x.a);}
    ModInt& operator-=(const ModInt &x){return s(a+mod-x.a);}
    ModInt& operator*=(const ModInt &x){
        a=uint64_t(a)*x.a%mod;
        return *this;
    }
    ModInt& operator/=(const ModInt &x){
        *this*=x.inv();
        return *this;
    }
 
    ModInt operator+(const ModInt &x)const{return ModInt(*this)+=x;}
    ModInt operator-(const ModInt &x)const{return ModInt(*this)-=x;}
    ModInt operator*(const ModInt &x)const{return ModInt(*this)*=x;}
    ModInt operator/(const ModInt &x)const{return ModInt(*this)/=x;}
    bool operator==(const ModInt &x)const{return a==x.a;}
    bool operator!=(const ModInt &x)const{return a!=x.a;}
    bool operator<(const ModInt &x)const{return a<x.a;}
 
    ModInt operator-()const{return ModInt()-*this;}
    ModInt pow(int64_t n)const{
        ModInt res(1),x(*this);
        while(n){
            if(n&1)res*=x;
            x*=x;
            n>>=1;
        }
        return res;
    }
 
    ModInt inv()const{return pow(mod-2);}
};
 
template<uint32_t mod>
istream& operator>>(istream& in,ModInt<mod>& a){
    return (in>>a.a);
}
template<uint32_t mod>
ostream& operator<<(ostream& out,const ModInt<mod>& a){
    return (out<<a.a);
}
using mint=ModInt<MOD>;

int main(void)
{
    int N;
    cin >> N;
    map<int, int> mp;
    REP(i, N){
        int a;
        cin >> a;
        mp[a]++;
    }

    if(N == 1){
        if(mp[0] == 1) pr(1);
        else pr(0);
        return 0;
    }
    
    if(N&1){
        if(mp[0] != 1){
            pr(0);
            return 0;
        }
        for (int i = 2; i < N; i+=2)
        {
            if(mp[i] != 2){
                pr(0);
                return 0;
            }
        }
        mint ans(2);
        REP(i, N/2-1){
            ans *= 2;
        }
        pr(ans);
    }else{
        for (int i = 1; i < N; i+=2)
        {
            if(mp[i] != 2){
                pr(0);
                return 0;
            }
        }
        mint ans(2);
        REP(i, N/2-1){
            ans *= 2;
        }
        pr(ans);
    }
}