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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
int main(){
  long long int x[5],b[100010],k=1,l=0,i,j,n,a[100010],mod=1000000007,v,ans=1;
  scanf("%lld\n",&n);
  for(i=n-1;i>=0;i--){
    scanf("%lld",&a[i]);
  }
  b[0]=3;
  x[0]=a[0];
  x[1]=-10,x[2]=-10;
  for(i=1;i<n;i++){
    if(a[i-1]<=a[i]){
      if(k<3){
        x[k]=a[i];
        b[i]=3-k;
        k++;
      }
      else{
        for(j=0;j<3;j++){
          if(x[j]-1==a[i]){
            v=j;
            l++;
          }
        }
        b[i]=l;
        x[v]--;
        l=0;
      }
    }
    else{
      for(j=0;j<3;j++){
        if(x[j]-1==a[i]){
          v=j;
          l++;
        }
      }
      b[i]=l;
      x[v]--;
      l=0;
    }
  }
  for(i=0;i<n;i++){
    ans=ans*b[i];
    ans=ans%mod;
  }
  printf("%lld\n",ans);
  return 0;
}