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: Accepted)

#include <stdio.h>

int main(void){

  int n;
  scanf("%d", &n);
	int a[n];
	for (int i = 0; i < n; i++) {
		scanf("%d", &a[i]);
	}
	long mod = 1000000007;
	long count[n];
	for (int i = 0; i < n; i++) {
		count[i] = 0;
	}
	long answer;
	for (int i = 0; i < n; i++) {
		if (a[i] == 0) {
			if (count[0] == 0) {
				answer = 3;
			} else if (count[0] == 1) {
				answer *= 2;
			} else if (count[0] > 2) {
				printf("0\n");
				return 0;
			}
			count[0]++;
		} else {
			if (count[a[i]] == 0) {
				if (count[a[i]-1] == 0) {
					printf("0\n");
					return 0;
				}
				answer *= count[a[i]-1];
			} else if (count[a[i]] == 1) {
				if (count[a[i]-1] <= 1) {
					printf("0\n");
					return 0;
				}
				answer *= count[a[i]-1]-1;
			} else if (count[a[i]] == 2){
				if (count[a[i]-1] <= 2) {
					printf("0\n");
					return 0;
				}
			} else {
				printf("0\n");
				return 0;
			}
			count[a[i]]++;
			answer %= mod;
		}
	}
	printf("%ld\n", answer);

  return 0;
}