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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;

public class Program
{
    private int N;
    private int[] A;

    public void Solve()
    {
        var sc = new Scanner();
        N = sc.NextInt();
        A = sc.IntArray();
        var c = new int[3];
        ModInt ans = 1;
        foreach (int i in A)
        {
            int cnt = 0;
            int ind = -1;
            for (int j = 0; j < 3; j++)
            {
                if (c[j] == i)
                {
                    cnt++;

                    ind = j;
                }
            }

            ans *= cnt;
            if (cnt > 0)
                c[ind]++;
            else
                break;
        }

        Console.WriteLine(ans);
    }

    public static void Main(string[] args)
    {
        new Program().Solve();
    }
}

#region ModInt

/// <summary>
/// [0,<see cref="Mod"/>) までの値を取るような数
/// </summary>
public struct ModInt
{
    /// <summary>
    /// 剰余を取る値．
    /// </summary>
    public const long Mod = (int) 1e9 + 7;

    /// <summary>
    /// 実際の数値．
    /// </summary>
    public long num;

    /// <summary>
    /// 値が <paramref name="n"/> であるようなインスタンスを構築します．
    /// </summary>
    /// <param name="n">インスタンスが持つ値</param>
    /// <remarks>パフォーマンスの問題上，コンストラクタ内では剰余を取りません．そのため，<paramref name="n"/> ∈ [0,<see cref="Mod"/>) を満たすような <paramref name="n"/> を渡してください．このコンストラクタは O(1) で実行されます．</remarks>
    public ModInt(long n)
    {
        num = n;
    }

    /// <summary>
    /// このインスタンスの数値を文字列に変換します．
    /// </summary>
    /// <returns>[0,<see cref="Mod"/>) の範囲内の整数を 10 進表記したもの．</returns>
    public override string ToString()
    {
        return num.ToString();
    }

    public static ModInt operator +(ModInt l, ModInt r)
    {
        l.num += r.num;
        if (l.num >= Mod) l.num -= Mod;
        return l;
    }

    public static ModInt operator -(ModInt l, ModInt r)
    {
        l.num -= r.num;
        if (l.num < 0) l.num += Mod;
        return l;
    }

    public static ModInt operator *(ModInt l, ModInt r)
    {
        return new ModInt(l.num * r.num % Mod);
    }

    public static implicit operator ModInt(long n)
    {
        n %= Mod;
        if (n < 0) n += Mod;
        return new ModInt(n);
    }

    /// <summary>
    /// 与えられた 2 つの数値からべき剰余を計算します．
    /// </summary>
    /// <param name="v">べき乗の底</param>
    /// <param name="k">べき指数</param>
    /// <returns>繰り返し二乗法により O(N log N) で実行されます．</returns>
    public static ModInt Pow(ModInt v, long k)
    {
        return Pow(v.num, k);
    }

    /// <summary>
    /// 与えられた 2 つの数値からべき剰余を計算します．
    /// </summary>
    /// <param name="v">べき乗の底</param>
    /// <param name="k">べき指数</param>
    /// <returns>繰り返し二乗法により O(N log N) で実行されます．</returns>
    public static ModInt Pow(long v, long k)
    {
        long ret = 1;
        for (k %= Mod - 1; k > 0; k >>= 1, v = v * v % Mod)
            if ((k & 1) == 1)
                ret = ret * v % Mod;
        return new ModInt(ret);
    }

    /// <summary>
    /// 与えられた数の逆元を計算します．
    /// </summary>
    /// <param name="v">逆元を取る対象となる数</param>
    /// <returns>逆元となるような値</returns>
    /// <remarks>法が素数であることを仮定して，フェルマーの小定理に従って逆元を O(log N) で計算します．</remarks>
    public static ModInt Inverse(ModInt v)
    {
        return Pow(v, Mod - 2);
    }
}

#endregion

class Scanner
{
    public Scanner()
    {
        _pos = 0;
        _line = new string[0];
    }

    const char Separator = ' ';
    private int _pos;
    private string[] _line;

    #region スペース区切りで取得

    public string Next()
    {
        if (_pos >= _line.Length)
        {
            _line = Console.ReadLine().Split(Separator);
            _pos = 0;
        }

        return _line[_pos++];
    }

    public int NextInt()
    {
        return int.Parse(Next());
    }

    public long NextLong()
    {
        return long.Parse(Next());
    }

    public double NextDouble()
    {
        return double.Parse(Next());
    }

    #endregion

    #region 型変換

    private int[] ToIntArray(string[] array)
    {
        var result = new int[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = int.Parse(array[i]);
        }

        return result;
    }

    private long[] ToLongArray(string[] array)
    {
        var result = new long[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = long.Parse(array[i]);
        }

        return result;
    }

    private double[] ToDoubleArray(string[] array)
    {
        var result = new double[array.Length];
        for (int i = 0; i < array.Length; i++)
        {
            result[i] = double.Parse(array[i]);
        }

        return result;
    }

    #endregion

    #region 配列取得

    public string[] Array()
    {
        if (_pos >= _line.Length)
            _line = Console.ReadLine().Split(Separator);

        _pos = _line.Length;
        return _line;
    }

    public int[] IntArray()
    {
        return ToIntArray(Array());
    }

    public long[] LongArray()
    {
        return ToLongArray(Array());
    }

    public double[] DoubleArray()
    {
        return ToDoubleArray(Array());
    }

    #endregion
}