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 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
const long long int mod=1e9+7;
int a[100005];
long long int dp[100005];
int n;
void solve(int now,int x,int y,int z){
    if(now==n)return;
    else if((x==y && y==z) && z==(a[now]-1)){
        dp[now]=dp[now-1]*3;
        dp[now]%=mod;
        solve(now+1,x+1,y,z);
    }
    else if(x==y && y==(a[now]-1)){
        dp[now]=dp[now-1]*2;
        dp[now]%=mod;
        solve(now+1,x+1,y,z);        
    }
    else if(z==y && y==(a[now]-1)){
        dp[now]=dp[now-1]*2;
        dp[now]%=mod;
        solve(now+1,x,y+1,z);        
    }
    else if(z==x && x==(a[now]-1)){
        dp[now]=dp[now-1]*2;
        dp[now]%=mod;
        solve(now+1,x+1,y,z);        
    }
    else if(x==(a[now]-1)){
        dp[now]=dp[now-1];
        solve(now+1,x+1,y,z);
    }
    else if(y==(a[now]-1)){
        dp[now]=dp[now-1];
        solve(now+1,x,y+1,z);
    }
    else if(z==(a[now]-1)){
        dp[now]=dp[now-1];
        solve(now+1,x,y,z+1);
    }
}

int main(){
    cin >> n;
    for(int i=0;i<n;i++)cin >> a[i];
    dp[0]=3;
    solve(1,0,-1,-1);
    cout << dp[n-1];
}