Problem:
Score: 500 points

Problem Statement

N people are standing in a queue, numbered 1, 2, 3, ..., N from front to back. Each person wears a hat, which is red, blue, or green.

The person numbered i says:

"In front of me, exactly A_i people are wearing hats with the same color as mine."

Assuming that all these statements are correct, find the number of possible combinations of colors of the N people's hats.

Since the count can be enormous, compute it modulo 1000000007.

Constraints

1 \leq N \leq 100000

0 \leq A_i \leq N-1

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 A_3 ... A_N

Output

Print the number of possible combinations of colors of the N people's hats, modulo 1000000007.

Sample Input 1

6
0 1 2 3 4 5

Sample Output 1

3

We have three possible combinations, as follows:

Red, Red, Red, Red, Red, Red

Blue, Blue, Blue, Blue, Blue, Blue

Green, Green, Green, Green, Green, Green

Sample Input 2

3
0 0 0

Sample Output 2

6

Sample Input 3

54
0 0 1 0 1 2 1 2 3 2 3 3 4 4 5 4 6 5 7 8 5 6 6 7 7 8 8 9 9 10 10 11 9 12 10 13 14 11 11 12 12 13 13 14 14 15 15 15 16 16 16 17 17 17

Sample Output 3

115295190

Solution in Rust (1.15.1) (judge verdict: Accepted)

fn read1<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

pub fn readn<T: std::str::FromStr>(delimiter: &str) -> Vec<T> {
    let s = read1::<String>();
    s.split(delimiter).map(|e| e.parse().ok().unwrap()).collect::<Vec<T>>()
}

const MOD: usize = 1_000_000_007;

fn main() {
    let n = read1::<usize>();
    let a = readn::<usize>(" ");
    let mut c = vec![1, 0, 0];
    let mut dp = vec![0; n];
    dp[0] = 3;

    if a[0] != 0 {
        println!("0");
        return;
    }

    for i in 1..n {
        let mut k = 0;
        for j in 0..3 {
            if c[j] == a[i] {
                if k == 0 {
                    c[j] += 1;
                }
                k += 1;
            }
        }
        if k == 0 {
            println!("0");
            return;
        }
        dp[i] = dp[i-1] * k % MOD;
    }

    println!("{}", dp[n-1]);
}