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: Accepted)

//   _/                _/                 _/_/_/   _/
//_/_/_/_/   _/_/   _/_/_/_/   _/_/    _/       _/_/
// _/     _/    _/   _/     _/    _/  _/_/_/     _/
//_/     _/    _/   _/     _/    _/  _/    _/   _/
// _/_/   _/_/       _/_/   _/_/      _/_/     _/
#include<iostream>
#include<algorithm>
#include<cmath>
#include<iomanip>
#include<set>
#include<map>
#include<queue>
#include<vector>
using namespace std;
typedef long long ll;
const int MOD=1e9+7;
const double pi=acos(-1);
int main() {
  cin.tie(0),cout.tie(0);
  ios::sync_with_stdio(false);
  ll n,a[100005]={},c[3]={},ans=1;
  cin >> n;
  for(int i=0; i<n; i++) {
    cin >> a[i];
  }
  for(int i=0; i<n; i++) {
    ll cnt=0,x=-1;
    if(a[i]==c[0]) {
      cnt++,x=0;
    }
    if(a[i]==c[1]) {
      cnt++,x=1;
    }
    if(a[i]==c[2]) {
      cnt++,x=2;
    }
    if(x==-1) {
      cout << 0 << endl;
      return 0;
    }
    c[x]++;
    ans*=cnt;
    ans%=MOD;
  }
  cout << ans << endl;
}