Codeforces-698A Vacations

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.

题目大意：

给你n天，每天能做的事件为ai，其中ai为0表示只能休息，ai为1表示可以休息或者打比赛，ai为2表示可以休息或者运动，ai为3表示可以休息，打比赛和运动

其中不可以连续两天打比赛或者连续两天运动

然后问你n天里面最少的休息时间。

解题思路：

DP

dp[i][j]表示第i天去做第j件事的时候最小的休息时间，j可以为0（表示休息），1（表示打比赛），2（表示运动）

代码：

#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;

#define mod 7
#define INF 0x3f3f3f3f
#define lson (rt << 1)
#define rson (rt << 1 | 1)
#define Clear(a) memset(a, 0, sizeof(a))
#define mp(a, b) make_pair((a), (b))
#define Max(a, b) ( (a) > (b) ? (a) : (b) )
#define Min(a, b) ( (a) < (b) ? (a) : (b) )

typedef long long LL;
typedef pair<int, int > pi;

const int maxn = 1e2 + 5;
const int dir[8][2] = {1,2, 2,1, -1,2, -2,1, 1,-2, 2,-1, -1,-2, -2,-1};

int a[maxn], dp[maxn][4];
int main()
{
    int n, val;
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i){
        scanf("%d", &a[i]);
    }
    memset(dp, 0, sizeof(dp));
    for(int i = 1; i <= n; ++i){
        if(a[i] == 0){
            dp[i][0] = min(dp[i-1][0], min(dp[i-1][1], dp[i-1][2])) + 1;
            dp[i][1] = dp[i][2] = INF;
        }else if(a[i] == 1){
            dp[i][0] = min(dp[i-1][0], min(dp[i-1][1], dp[i-1][2])) + 1;
            dp[i][1] = min(dp[i-1][0], dp[i-1][2]);
            dp[i][2] = INF;
        }else if(a[i] == 2){
            dp[i][0] = min(dp[i-1][0], min(dp[i-1][1], dp[i-1][2])) + 1;
            dp[i][1] = INF;
            dp[i][2] = min(dp[i-1][0], dp[i-1][1]);;
        }else if(a[i] == 3){
            dp[i][0] = min(dp[i-1][0], min(dp[i-1][1], dp[i-1][2])) + 1;
            dp[i][1] = min(dp[i-1][0], dp[i-1][2]);;
            dp[i][2] = min(dp[i-1][0], dp[i-1][1]);;
        }
    }
    printf("%d
", min(dp[n][0], min(dp[n][1], dp[n][2])));
    return 0;
}