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: Accepted)

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
ll MOD = 1000000007;

#define vec              vector<int>
#define vecll            vector<ll>
#define vecd             vector<double>
#define vecst            vector<string>
#define vecb             vector<bool>
#define v2(v,n,m,init)   vector<vector<int>>  v(n, vector<int>(m, init))
#define vb2(v,n,m,init)  vector<vector<bool>> v(n, vector<bool>(m, init))
#define vll2(v,n,m,init) vector<vector<ll>>   v(n, vector<ll>(m, init))

#define rep(i,n)         for(ll i=(ll)0; i<(ll)n; i++)
#define REP(i,m,n)       for(ll i=(ll)m; i<(ll)n; i++)
#define arr(var, n)      vec var(n); rep(i,n){cin >> var[i];}
#define arrll(var, n)    vecll var(n); rep(i,n){cin >> var[i];}
#define arrst(var, n)    vecst var(n); rep(i,n){cin >> var[i];}
#define all(var)         (var).begin(), (var).end()
#define sortall(var)     sort(all(var))
#define uniqueall(v)     v.erase(unique(v.begin(), v.end()), v.end());
#define pb(var)          push_back(var)
#define prt(var)         cout << (var) << "\n"
#define prtd(n, var)     cout << fixed << setprecision(n) << (var) << "\n"
#define prtfill(n, var)  cout << setw(n) << setfill('0') << (var);
#define prt2(v1, v2)     cout << (v1) << " " << (v2) << "\n"
#define prt3(v1, v2, v3) cout << (v1) << " " << (v2) << " " << (v3) << "\n"
#define prtall(v)        rep(i,v.size()){cout<<v[i]<<(i!=v.size()-1?" ":"\n");}
template <typename T>
bool chmax(T &a, const T& b){if(a<b){a=b; return true;} return false;}
template <typename T>
bool chmin(T &a, const T& b){if(a>b){a=b; return true;} return false;}
//------------------------------------------------------------------

int main(void){
  int n;
  cin >> n;
  arr(a,n);

  vec ct(3);
  ll ans=1;
  rep(i,n){
    ll tmp=0;
    bool already=false;
    rep(j,3){
      if(ct[j]==a[i]){
        tmp++;
        if(!already) ct[j]++, already=true;
      }
    }
    ans *= tmp;
    ans %= 1000000007;
  }

  prt(ans);
}