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: Runtime Error)

#include <bits/stdc++.h>
#define rep(i,n) for(int i = 0; i < (int)(n); i++)
#define fi first
#define se second
#define pb push_back
#define eb emplace_back
#define sz(x) (int)(x).size()
using namespace std;
using ll = long long;
using P = pair<int, int>;
using vi = vector<int>;
using vc = vector<char>;
using vb = vector<bool>;
using vs = vector<string>;
using vll = vector<long long>;
using vp = vector<pair<int, int>>;
using vvi = vector<vector<int>>;
using vvc = vector<vector<char>>;
using vvll = vector<vector<long long>>;
template<class T> inline bool chmax(T &a, T b) { if (a<b) {a=b; return 1;} return 0;}
template<class T> inline bool chmin(T &a, T b) { if (b<a) {a=b; return 1;} return 0;}

const int mod = 1000000007;
// const int mod = 998244353;
struct mint {
    long long x;
    mint(long long x=0) : x((x % mod + mod) % mod) {}
    mint operator-() const { return mint(-x);}
    mint& operator+=(const mint a) {
        if ((x += a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator-=(const mint a) {
        if ((x += mod - a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
    mint operator+(const mint a) const { return mint(*this) += a;}
    mint operator-(const mint a) const { return mint(*this) -= a;}
    mint operator*(const mint a) const { return mint(*this) *= a;}
    mint pow(long long t) const {
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }
    mint inv() const { return pow(mod-2);}
    mint& operator/=(const mint a) { return *this *= a.inv();}
    mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    int n;
    cin >> n;
    vi a(n);
    rep(i, n) cin >> a[i];
    vi cnt(n-1, 0);

    mint ans = 1;
    rep(i, n) {
        if (a[i] == 0) {
            ans *= 3 - cnt[0];
            cnt[0]++;
        } else {
            ans *= cnt[a[i] - 1] - cnt[a[i]];
            cnt[a[i]]++;
        }
    }
    cout << ans << endl;
}