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 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main()
{
	int i, N, A[100001];
	scanf("%d", &N);
	for (i = 1; i <= N; i++) scanf("%d", &(A[i]));
	
	int j, count[5] = {1, 0, 0, -1, -1};
	long long ans = 3;
	for (i = 2; i <= N; i++) {
		for (j = 0; j < 3; j++) {
			if (A[i] == count[j]) break;
		}
		// if (j == 3) ans *= 0;
		// else {
			if (count[j+1] == count[j]) {
				if (count[j+2] != count[j]) ans = ans * 2 % 1000000007;
				else ans = ans * 3 % 1000000007;
			}
			count[j]++;
		// }
	}
	
	printf("%lld\n", ans);
	fflush(stdout);
	return 0;
}