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 PHP7 (7.0.15) (judge verdict: Accepted)

<?php
define("MOD", 10 ** 9 + 7);
list($n) = ints();
$a = ints();
for ($i = 0; $i < $n; ++$i) {
    if (isset($c[$a[$i]])) $c[$a[$i]]++;
    else $c[$a[$i]] = 1;
}
if ($n % 2 && $c[0] !== 1) exit('0');
for ($i = 0; $i < intdiv($n, 2); ++$i) {
    $k = abs($i - ($n - $i - 1));
    if ($c[$k] !== 2) exit('0');
}
echo modPow(2, intdiv($n, 2));
function ints()
{
    return array_map('intval', explode(' ', trim(fgets(STDIN))));
}

// 足し算
function modAdd($x, $y)
{
    return ($x + $y) % MOD;
}

// 引き算
function modSub($x, $y)
{
    return ($x + MOD - $y) % MOD;
}

// 掛け算
function modMul($x, $y)
{
    return ($x * $y) % MOD;
}

// 割り算
function modDiv($x, $y)
{
    return modMul($x, modPow($y, MOD - 2));
}

// 累乗（繰り返し二乗法）
function modPow($n, $x)
{
    if ($x === 0) return 1;
    $res = (modPow($n, $x >> 1) ** 2) % MOD;
    if ($x % 2 === 1) $res = modMul($res, $n);
    return $res;
}