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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
import java.io.*;

public class Main {
	static int mod = (int) 1e9 + 7;
	static int DX[] = { -1, 0, 1, 0 }, DY[] = { 0, -1, 0, 1 };

	public static void main(String[] args) {
		FastScanner fs = new FastScanner(System.in);
		int n = fs.nextInt();
		int a[] = fs.nextIntArray(n);
		Arrays.sort(a);

		int memo[] = new int[n];
		for(int i=0;i<n;i++) {
			memo[a[i]]++;
		}
		
		if(n%2==0) {
			for(int i=1;i<n;i+=2) {
				if(memo[i]!=2) {
					System.out.println(0);
					return;
				}
			}
		}
		else {
			for(int i=0;i<n;i+=2) {
				if(i==0) {
					if(memo[i]!=1) {
						System.out.println(0);
						return;
					}
				}
				else {
					if(memo[i]!=2) {
						System.out.println(0);
						return;
					}
				}
			}
		}
		
		long ans = modpow(2,n/2);
		System.out.println(ans);
		
	}
	//MOD culculations
	static long plus(long x, long y) {
		x %= mod;
		y %= mod;
		long res = (x + y) % mod;
		return res;
	}

	static long sub(long x, long y) {
		x %= mod;
		y %= mod;
		long res = (x - y + mod) % mod;
		return res;
	}

	static long mul(long x, long y) {
		x %= mod;
		y %= mod;
		long res = x * y % mod;
		return res;
	}

	static long div(long x, long y) {
		x %= mod;
		y %= mod;
		long res = x * modpow(y, mod - 2) % mod;
		return res;
	}

	//modの元での高速pow()
	static long modpow(long a, long p) {
		if (p == 0)
			return 1;
		if (p % 2 == 0) {
			long halfP = p / 2;
			long root = modpow(a, halfP);
			return root * root % mod;
		} else
			return a * modpow(a, p - 1) % mod;
	}

}

//高速なScanner
class FastScanner {
	private BufferedReader reader = null;
	private StringTokenizer tokenizer = null;

	public FastScanner(InputStream in) {
		reader = new BufferedReader(new InputStreamReader(in));
		tokenizer = null;
	}

	public String next() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				tokenizer = new StringTokenizer(reader.readLine());
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken();
	}

	public String nextLine() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				return reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken("\n");
	}

	public long nextLong() {
		return Long.parseLong(next());
	}

	public int nextInt() {
		return Integer.parseInt(next());
	}

	public double nextDouble() {
		return Double.parseDouble(next());
	}

	public int[] nextIntArray(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = nextInt();
		return a;
	}

	public long[] nextLongArray(int n) {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nextLong();
		return a;
	}
}