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++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <cmath>
#include <set>
#include <algorithm>

typedef long long ll;
const ll MOD = 1e9 + 7;
const long long INF = 1LL<<60;
const double PI=3.14159265358979323846;
const int NMAX=100005;
using namespace std;

vector<int> a(100002);

int main(){
    ll n;
    cin >> n;
    for(int i = 0; i < n; i++){
        cin >> a[i];
    }
    
    vector<int> c;
    ll ans = 1;
    for(int i = 0; i < n; i++){
        if(a[i] == 0){
            c.push_back(0);
        } else {
            ll pos = 0;
            int tmp = 0;
            for(int j = 0; j < c.size(); j++){
                if(c[j] == a[i]-1){
                    tmp = j;
                    pos++;
                }
            }
            c[tmp]++;
            ans = ans*pos %MOD;
        }
    }

    for(ll j = 0; j < c.size(); j++){
        ans = ans * (3L-j) % MOD;
    }

    cout << ans << endl;
    return 0;
}