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: Accepted)

// Generated by 1.1.6 https://github.com/kyuridenamida/atcoder-tools

#include <bits/stdc++.h>

using namespace std;

using i64 = long;

const long long MOD = 1000000007;
const i64 INF = i64(1e18) + 7;


template <typename T>
bool chmin(T& x, T y){
    if(x > y){
        x = y;
        return true;
    }
    return false;
}

template <typename T>
bool chmax(T& x, T y){
    if(x < y){
        x = y;
        return true;
    }
    return false;
}

template <i64 mod = MOD>
struct ModInt{
    i64 p;

    ModInt() : p(0){}
    ModInt(i64 x){p = x >= 0 ? x % mod : x + (-x + mod - 1) / mod * mod;}

    ModInt& operator+=(const ModInt& y){p = p + *y - ((p + *y) >= mod ? mod : 0); return *this;}
    ModInt& operator-=(const ModInt& y){p = p - *y + (p - *y < 0 ? mod : 0); return *this;}
    ModInt& operator*=(const ModInt& y){p = (p * *y) % mod; return *this;}
    ModInt& operator%=(const ModInt& y){if(y)p %= *y; return *this;}

    ModInt operator+(const ModInt& y) const{ModInt x = *this; return x += y;}
    ModInt operator-(const ModInt& y) const{ModInt x = *this; return x -= y;}
    ModInt operator*(const ModInt& y) const{ModInt x = *this; return x *= y;}
    ModInt operator%(const ModInt& y) const{ModInt x = *this; return x %= y;}

    friend ostream& operator<<(ostream& stream, const ModInt<mod>& x){
        stream << *x;
        return stream;
    }

    friend ostream& operator>>(ostream& stream, const ModInt<mod>& x){
        stream >> *x;
        return stream;
    }

    ModInt& operator++(){p = (p + 1) % mod; return *this;}
    ModInt& operator--(){p = (p - 1 + mod) % mod; return *this;}

    bool operator==(const ModInt& y) const{return p == *y;}
    bool operator!=(const ModInt& y) const{return p != *y;}

    const i64& operator*() const{return p;}
    i64& operator*(){return p;}

};

using mint = ModInt<>;


void solve(long long n, std::vector<long long> a){
    vector<int> cnt(n + 1, 0);
    cnt[0] = 3;
    mint ans = 1;
    for(auto& x : a){
        if(cnt[x] == 0){
            ans = 0;
            break;
        }
        ans *= cnt[x];
        --cnt[x];
        ++cnt[x + 1];
    }
    cout << *ans << endl;
}


signed main(){
    long long N;
    scanf("%lld",&N);
    std::vector<long long> A(N);
    for(int i = 0 ; i < N ; i++){
        scanf("%lld",&A[i]);
    }
    solve(N, std::move(A));
    return 0;
}