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 <limits.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#include <math.h>

#define P 1000000007

int comp(const void *a, const void *b){return *(int*)a-*(int*)b;}
int compw(const void *a, const void *b){return (*(int*)a>*(int*)b)-(*(int*)a<*(int*)b);}
int compr(const void *a, const void *b){return *(int*)b-*(int*)a;}
int compl(const void *a, const void *b){return (*(long*)a>*(long*)b)-(*(long*)a<*(long*)b);}

uint32_t nextpint(void){ char c=getchar(); while(c<'0'||'9'<c) c=getchar(); uint_fast32_t x=0; while('0'<=c && c<='9'){ x=x*10+c-'0'; c=getchar(); } return x; }

inline int min2(const int a, const int b){return a>b?b:a;}
inline int dif2(const int a, const int b){return a>b?a-b:b-a;}

int s[3];
int main(void){
	int n=nextpint();
	long ans=1;
	for(int i=0; i<n; i++){
		int a=nextpint();
		int b=(s[0]==a)+(s[1]==a)+(s[2]==a);
		ans=(ans*b)%P;
		if(0);
		else if(s[0]==a) s[0]++;
		else if(s[1]==a) s[1]++;
		else if(s[2]==a) s[2]++;
	}
	printf("%d\n", ans);
}