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 <fstream>
#include <iostream>
#include <functional>
#include <algorithm>
#include <sstream>
#include <cstring>
#include <vector>
#include <string>
#include <cstdio>
#include <cmath>
#include <queue>
#include <stack>
#include <deque>
#include <ctime>
#include <list>
#include <set>
#include <map>
#include <bitset>
using namespace std;
 
typedef long long ll;
typedef unsigned long long ull;
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define GET_POS(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())
#define CASET int ___T; scanf("%d", &___T); for(int cs=1;cs<=___T;cs++)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MS1(X) memset((X), -1, sizeof((X)))
#define EPS 1e-8
#define LL_INF 0x3fffffffffffffff
#define INF 0x3f3f3f3f3f3f3f3f
#define MEM(a, b) memset(a, b, sizeof(a))
#define PPER(i, n, m) for (int i = n; i >= m; i--)
#define REPP(i, n, m) for (int i = n; i <= m; i++)
#define REP(i, n, m) for (int i = n; i < m; i++)
#define PER(i, n, m) for (int i = n; i > m; i--)
#define SA(n) scanf("%d", &(n))
#define SLLA(n) scanf("%lld", &(n))
#define MP make_pair
#define FF first
#define SS second
#define PB push_back
#define DE(val) cout << #val << ": " << val << endl;

const int dx44[6] = {0, -1, -1, 1, 1};
const int dy44[6] = {0, -1, 1, -1, 1};
const int dx4[6] = {0, 1, 0,-1};
const int dy4[6] = {-1, 0, 1,0};
const int dx8[9] = {0, -1, 0, 1, 0, 1, 1, -1, -1};
const int dy8[9] = {0, 0, 1, 0, -1, 1, -1, 1, -1};
const int dx82[9] = {0, -1, -1, 1, 1, 2, 2, -2, -2};
const int dy82[9] = {0, 2, -2, 2, -2, 1, -1, 1, -1};
void update(ll& x,ll v){
    if(x==-1)x=v;
    else if(x>v)x=v;
}

ll po(ll a, ll b, ll mod)
{
    ll res = 1;
    a %= mod;
    for (; b; b >>= 1)
    {
        if (b & 1)
            res = res * a % mod;
        a = a * a % mod;
    }
    return res;
}
ll gcd(ll a, ll b)
{
    if (a == 0)
    {
        return b;
    }
    else
    {
        return gcd(b % a, a);
    }
}
void extgcd(ll a, ll b, ll &d, ll &x, ll &y)
{
    if (!b)
    {
        d = a;
        x = 1;
    }
    else
    {
        extgcd(b, a % b, d, y, x);
        y -= x * (a / b);
    }
}
ll inverse(ll a, ll n)
{
    ll d, x, y;
    extgcd(a, n, d, x, y);
    return d == 1 ? (x + n) % n : -1;
}
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
const int maxn = 3e5 + 5;
const double PI = acos(-1.0);
const ll mod = 1000000007;
/*************************************************************************/

int n;
int val[maxn];
void solve() {
    SA(n);
    REPP(i,1,n) {
        SA(val[i]);
    }

    ll ans = 3;
    map<int,ll> hax;
    hax[0] = 1;
    hax[1] = 0;
    hax[2] = 0;
    REPP(i,2,n) {

        ll cnt = 0;
        REPP(k,0,2) {
            if(val[i] == hax[k]) {
                cnt++;
            }
        }
        REPP(k,0,2) {
            if(val[i] == hax[k]) {
                hax[k]++;
                break;
            }
        }
        ans = ans*cnt % mod;
        if(ans == 0) {
            while(true);
        }
        ans%=mod;
    }
    printf("%lld\n", ans);
}

int main()
{
#ifndef ONLINE_JUDGE
 // freopen("i.txt", "r", stdin);
  //freopen("o.txt", "w", stdout);
#endif
  solve();

  return 0;
}