Vasya has n days of vacations! So he decided to imPRove his IT skills and do sport. Vasya knows the following information about each of this n days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the i-th day there are four options:

On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).

Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.

The first line contains a positive integer n (1?≤?n?≤?100) — the number of days of Vasya’s vacations.

The second line contains the sequence of integers a₁,?a₂,?…,?a_n (0?≤?a_i?≤?3) separated by space, where:

Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:

41 3 2 0Output
2Input
71 3 3 2 1 2 3Output
0Input
22 2Output
1Note
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
  
題意:: 一個人在一天內(nèi)有四種選擇 
休息，運動||休息，工作||休息，運動||工作 工作或者運動不能連續(xù)超過兩天，求最短的休息時間  重點就是不要連續(xù)超過兩天就可以 數(shù)據(jù)比較小，暴力就可以AC了
#include<stdio.h>int arr[105];int main(){    int n;    while(~scanf("%d",&n)) {        for(int i = 0;i < n; i++)             scanf("%d",&arr[i]);            int k = 0, flag = 0;        for(int i = 0;i < n; i++) {            if(arr[i] == 0) {                k++;                flag = 0;            }            else if(arr[i]==1 || arr[i]==2) {                if(flag == arr[i]) {                    k++;                    flag = 0;                }                else                     flag =arr[i];            }            else {                if(flag == 1) {                    flag = 2;                }                else if(flag == 2) {                    flag = 1;                }            }        }        printf("%d/n",k);     }return 0;}