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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.*;
 
public class Main {
 
    void solve() throws IOException {
        int N = ni();
        int[] a = new int[N];
        for (int i = 0; i < N; i++) {
            a[i] = ni();
        }
        int[] cand = new int[N];
        cand[0] = 3;
        int ans = 1;
        for (int i = 0; i < a.length; i++) {
            ans = mul(ans, cand[a[i]]);
            cand[a[i]]--;
            if(a[i]+1<N-1) cand[a[i]+1]++;
        }
 
        out.println(ans);
    }

    final int mod = 1000000007;
    final BigInteger MOD = BigInteger.valueOf(mod);

    int mul(int x, int y){
        int val = (int)((x * 1L * y) % mod);
        return val>=0 ? val : val+mod;
    }
    int add(int x, int y) {
        x += y;
        if(x < 0) x += mod;
        if(x>=mod) x -= mod;
        return x;
    }
    int sub(int x, int y){
        x = add(x,mod-y);
        if(x < 0) x += mod;
        if(x>=mod) x -= mod;
        return x;
    }
    String ns() throws IOException {
        while (!tok.hasMoreTokens()) {
            tok = new StringTokenizer(in.readLine(), " ");
        }
        return tok.nextToken();
    }
 
    int ni() throws IOException {
        return Integer.parseInt(ns());
    }
 
    long nl() throws IOException {
        return Long.parseLong(ns());
    }
 
    double nd() throws IOException {
        return Double.parseDouble(ns());
    }
 
    String[] nsa(int n) throws IOException {
        String[] res = new String[n];
        for (int i = 0; i < n; i++) {
            res[i] = ns();
        }
        return res;
    }
 
    int[] nia(int n) throws IOException {
        int[] res = new int[n];
        for (int i = 0; i < n; i++) {
            res[i] = ni();
        }
        return res;
    }
 
    long[] nla(int n) throws IOException {
        long[] res = new long[n];
        for (int i = 0; i < n; i++) {
            res[i] = nl();
        }
        return res;
    }
 
    static BufferedReader in;
    static PrintWriter out;
    static StringTokenizer tok;
 
    public static void main(String[] args) throws IOException {
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        tok = new StringTokenizer("");
        Main main = new Main();
        main.solve();
        out.close();
    }
}