vacation ——————hdu codeforces 986a

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

题意分析：可以看成一个贪心的实现，跟之前做的那个电视台差不多（或者说是那个暑假不ac）只不过暑假不ac是对坐标的右边进行排序，而这个是对左边排序，小的在左面

这道题里的一个技巧是可能出现在最左面一个点是覆盖所有点的情况，于是就把每个点的右端点值进行比较

代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string.h>

using namespace std;

int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)&&(n||m))
    {
        n=n*100;
       // int a,b;
        int flag=0;
        for(int i=0;i<=9;i++)
        {
            for(int j=0;j<=9;j++)
            {
                int sum=i*10+j;
                if((n+sum)%m==0)
                {
                   if(flag==0)
                   {
                       flag=1;
                       printf("%d%d",i,j);
                   }
                   else printf(" %d%d",i,j);
                }
            }
        }
        puts("");
    }
    return  0;
}