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)

#pragma GCC optimize("Ofast")
#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define rep(i,n) for(int i=0;i<n;i++)
#define INF 1000000000000000000
#define MOD 1000000007
typedef pair<int, int> P;

bool prime(int n) {
	for (int i = 2; i <= sqrt(n); i++) {
		if (n % i == 0)return false;
	}
	return n != 1;
}
int gcd(int x, int y) {
	if (y == 0)return x;
	return gcd(y, x % y);
}
int lcm(int x, int y) {
	return x / gcd(x, y) * y;
}
int mod_pow(int n, int p, int m) {
	if (p == 0) return 1;
	if (p % 2 == 0) {
		int t = mod_pow(n, p / 2, m);
		return (t * t) % m;
	}
	return n * mod_pow(n, p - 1, m) % m;
}
int extGCD(int a, int b, int& x, int& y) {
	if (b == 0) {
		x = 1;
		y = 0;
		return a;
	}
	long long d = extGCD(b, a % b, y, x);
	y -= a / b * x;
	return d;
}
int modinv(int a, int m) {
	int b = m, u = 1, v = 0;
	while (b) {
		long long t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= m;
	if (u < 0) u += m;
	return u;
}
int digit(int x) {
	int cnt = 0;
	while (x > 0) {
		cnt += x % 10;
		x /= 10;
	}
	return cnt;
}
int read() {
	int tmp;
	cin >> tmp;
	return tmp;
}
int xor128() {
	static int x = 123456789, y = 362436069, z = 521288629, w = 88675123;
	int t = (x xor (x << 11));
	x = y; y = z; z = w;
	return (w = (w xor (w >> 19)) xor (t xor (t >> 8)));
}

signed main() {
	int n;
	cin >> n;
	int cnt[3] = { 0,0,0 };
	int ans = 1;
	rep(i, n) {
		int a;
		cin >> a;
		int pl = 0;
		rep(j, 3) {
			if (cnt[j] == a) pl++;
		}
		ans *= pl;
		ans %= MOD;
		rep(j, 3) {
			if (cnt[j] == a) {
				cnt[j]++;
				break;
			}
		}
	}
	cout << ans << endl;

	return 0;
}