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)

#define _GLIBCXX_DEBUG//TLEの原因になるので注意！！！！！！！！！！！
#include<cmath>
#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
vector<int> arr;
stack<int> st;
queue<int> qu;
queue<pair<int,int> > qu2;
priority_queue<int> pq;
#define rep(i,n) for(int i=0; i<(int)(n); i++)
#define rep2(i,n) for(int i=1; i<=(int)(n); i++)
#define mins(x,y) x=min(x,y)
#define maxs(x,y) x=max(x,y)
typedef set<int> set_t;
typedef set<string> set_g;
typedef complex<double> xy_t;
static const int NIL = -1;
static const int INF = 1000000007;
#define mp make_pair
#define sz(x) int(x.xise())
#define mod 1000000007
#define reps(i,s,n) for(int i = s; i < n; i++)
deque<int> deq;
//#define int long long

//cerrはコメントアウトしなくても通る//ただしTLEを引き起こすことがある

int a[100005];
int d[100005];
int used[100005];


int main(){
    int n;
    cin >> n;
    rep(i,n){
        cin >> a[i];
    }
    ll ans=1;
    rep(i,n){
        int num=a[i];
        if(num==0){
            if(d[0]==0){
                ans*=3;
            }else if(d[0]==1){
                ans*=2;
            }else if(d[0]==2){
                ans*=1;
            }else{
                cout<<0<<endl;
                return 0;
            }
            d[num]++;
            continue;
        }
        if(d[num-1]==0||d[num-1]>=4){
            cout<<0<<endl;
            return 0;
        }
        if(d[num-1]==1){
            ans*=1-used[num-1];
        }else if(d[num-1]==2){
            ans*=2-used[num-1];
        }else if(d[num-1]==3){
            ans*=3-used[num-1];
        }
        used[num-1]++;
        d[num]++;
    }
    ans%=mod;
    cout<<ans<<endl;
    return 0;
}