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>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <stdbool.h>
#define LL long long
#define INF 1<<31-1
#define LNF 9223372036854775807
#define PI 3.14159265358979
#define MIN(x,y) ((x)<(y)?(x):(y))
#define MAX(x,y) ((x)<(y)?(y):(x))
#define FOR(i,a,n) for(i=a;i<n;i++)
#define MOD 1000000007
#define NUM 100005

void swap(LL* a,LL* b){LL tmp=*a;*a=*b;*b=tmp;}
int abs(int a){if(a<0)return -a;return a;}
int dup(int a,int d){if(a%d==0)return a/d;else return a/d+1;}
int mod(int a,int m){if(a<0)return a%m+m;else return a%m;}
int GCD(int a,int b){int c=1,tmp=MAX(a,b);b=MIN(a,b);a=tmp;while(c!=0){c=a%b;a=b;b=c;}return a;}
int LCM(int a,int b){return a*b/GCD(a,b);}
int POW(int a,int n){LL ans=1;while(n>0){if(n&1)ans=ans*a%MOD;a=a*a%MOD;n>>=1;}return (int)ans;}
int icmp(const void* a,const void* b){return *(int*)a-*(int*)b;}
int lcmp(const void* a,const void* b){if(*(LL*)a>*(LL*)b)return 1;if(*(LL*)a<*(LL*)b)return -1;return 0;}
int ccmp(const void* a,const void* b){return *(char*)a-*(char*)b;}
int scmp(const void* a,const void* b){return strcmp(*(char**)a,*(char**)b);}
LL fac[NUM],finv[NUM],inv[NUM];
void comi(){int i;fac[0]=fac[1]=1;finv[0]=finv[1]=1;inv[1]=1;FOR(i,2,NUM){fac[i]=fac[i-1]*i%MOD;inv[i]=MOD-inv[MOD%i]*(MOD/i)%MOD;finv[i]=finv[i-1]*inv[i]%MOD;}}
int com(int n,int k){if(n<k||n<0||k<0)return 0;return fac[n]*(finv[k]*finv[n-k]%MOD)%MOD;}


int main(){
  LL N,x=0,d=0,n=0,i,ans=1;
  scanf("%lld",&N);
  LL A[N],a;
  FOR(i,0,N)scanf("%lld",&A[i]);
  FOR(i,0,N){
    a=0;
    if(A[i]==x)a++;
    if(A[i]==d)a++;
    if(A[i]==n)a++;
    if(A[i]==x)x++;
    else if(A[i]==d)d++;
    else n++;
    ans=(ans*a)%MOD;
  }

  printf("%lld\n",ans);
	return 0;
}