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 <iostream>
#include <vector>
#include <array>
#include <list>
#include <queue>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <string>
#include <sstream>
#include <algorithm>
#include <random>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <cmath>
#include <cassert>
#include <climits>
#include <bitset>
#include <functional>
#include <iomanip>

#define FOR_LT(i, beg, end) for (int i = beg; i < end; i++)
#define FOR_LE(i, beg, end) for (int i = beg; i <= end; i++)
#define FOR_DW(i, beg, end) for (int i = beg; end <= i; i--)

using namespace std;

int64_t kRem = 1000000007;

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout << fixed << setprecision(20);

	int n; cin >> n;
	int nums[3] = { -1, -1, -1 };

	int64_t ans = 1;
	FOR_LT(i, 0, n) {
		int num; cin >> num;
		int64_t c = 0;
		bool sumi = false;
		FOR_LT(j, 0, 3) {
			if (nums[j] == num - 1) {
				c++;
				if (!sumi) {
					nums[j]++;
					sumi = true;
				}
			}
		}
		assert(sumi);
		ans *= c;
		ans %= kRem;
	}

	cout << ans << endl;

	return 0;
}