#include <iostream>
#include <stdio.h>
#include <cstring>
#include <algorithm>
#include <vector>
#include <map>
using namespace std;
const long long mod = 1e9+7;
typedef long long int ll;
#define maxn 100+5
#define INF 0x3f3f3f3f
#define LLF 0x7fffffffffffffff
int a[maxn],dp[maxn][maxn];
int main(){
    // #ifndef ONLINE_JUDGE    //if not define 如果没有定义这个的话就执行下面
	// freopen("input.in", "r", stdin);   //只改变输入流的文件指针，读入这个文件的内容（必须要有input这个文件）stdin是标准输入流的文件指针
	// freopen("output1.out", "w", stdout);  //只改变输出流的文件指针，写入output内（如果没有output这个文件就会自动生成）stdout是标准输出流的文件指针
	// #endif
    int n;
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>a[i];
    }
    memset(dp,100,sizeof(dp));
    dp[0][0]=0;
    for(int i=1;i<=n;i++){
        dp[i][0]=min(dp[i-1][1],min(dp[i-1][2],dp[i-1][0]))+1;
        if(a[i]==2||a[i]==3)
        dp[i][1]=min(dp[i-1][0],dp[i-1][2]);
        if(a[i]==1||a[i]==3)
        dp[i][2]=min(dp[i-1][1],dp[i-1][0]);
        
    }
    cout<<min(min(dp[n][1],dp[n][2]),dp[n][0])<<endl;
    
}

D - Vacations

 CodeForces - 699C 

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:

  1. on this day the gym is closed and the contest is not carried out;
  2. on this day the gym is closed and the contest is carried out;
  3. on this day the gym is open and the contest is not carried out;
  4. on this day the gym is open and the contest is carried out.

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.

Input

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, ..., an (0 ≤ ai ≤ 3) separated by space, where:

  • ai equals 0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
  • ai equals 1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
  • ai equals 2, if on the i-th day of vacations the gym is open and the contest is not carried out;
  • ai equals 3, if on the i-th day of vacations the gym is open and the contest is carried out.

Output

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

  • to do sport on any two consecutive days,
  • to write the contest on any two consecutive days.

Examples

Input

4
1 3 2 0

Output

2

Input

7
1 3 3 2 1 2 3

Output

0

Input

2
2 2

Output

1

Note

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.

题意：相邻两天不能做一样的事情，问你最少休息多少天

题解：当前天数是上一天不同事情的的状态,写状态转移方程的时候没想明白不能去怎么办。。。后来明白是直接初始化为一个无穷大的就可以了，不能去就直接不用管