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: Wrong Answer)

#include <bits/stdc++.h>
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
using namespace std;
typedef long long   ll;
typedef unsigned long long   ull;
typedef long double ld;
typedef pair<ll, ll>    pll;
typedef pair<ull, ull>    pull;
typedef pair<int, int>  pii;
typedef pair<ld, ld>  pld;
ll n;
ll a[200000];
ll dp[200000];
ll cnt[200000];
ll mod = 1000000007;
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    
    cin >> n;
    for(ll i = 0; i < n; ++i)
        cin >> a[i];
    dp[0] = 3;
    cnt[0] = 2;
    cnt[1] = 1;
    for(ll i = 1; i < n; ++i){
        if(cnt[a[i]] == 0){
            cout << "0\n";
            return 0;
        }
        dp[i] = dp[i-1]*cnt[a[i]]%mod;
        --cnt[a[i]];
        ++cnt[a[i]+1];
    }
    cout << dp[n-1]%mod << "\n";
}