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.
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import static java.lang.System.out;
import java.util.*;
import java.io.*;
import java.math.*;
/*
-> Give your 100%, that's it!
-> Feynman Rule To Solve Any Problem:
    1. Read the problem.
    2. Think About It.
    3. Solve it!
*/

public class Template {
    
	static int mod = 1000000007;

	public static void main(String[] args){
        FastScanner sc = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
        int yo = 1;
        while (yo-- > 0) {
            int n = sc.nextInt();
            int[] a = sc.readInts(n);
            int[][] dp = new int[n][2];
            int count = 0;
            for(int i = 0; i < n; i++){
                if(a[i] == 0){
                    count++;
                }
                else if(a[i] == 1){
                    if(i != 0){
                        if(dp[i-1][0] == 1){
                            count++;
                        }
                        else {
                            dp[i][0] = 1;
                        }
                    }
                    else{
                        dp[i][0] = 1;
                    }
                }
                else if(a[i] == 2){
                    if(i != 0){
                        if(dp[i-1][1] == 1){
                            count++;
                        }
                        else {
                            dp[i][1] = 1;
                        }
                    }
                    else {
                        dp[i][1] = 1;
                    }
                }
                else if(a[i] == 3){
                    if(i != 0){
                        if(dp[i-1][1] == 1){
                            dp[i][0] = 1;
                        }
                        if(dp[i-1][0] == 1){
                            dp[i][1] = 1;
                        }
                    }
                }
            }
            out.println(count);
        }
        out.close();    
	}
    
   
    
    
    /*
         Random stuff to try when stuck:
            -if it's 2C then it's dp
            -for combo/probability problems, expand the given form we're interested in
            -make everything the same then build an answer (constructive, make everything 0 then do something)
            -something appears in parts of 2 --> model as graph
            -assume a greedy then try to show why it works
            -find way to simplify into one variable if multiple exist
            -treat it like fmc (note any passing thoughts/algo that could be used so you can revisit them)
            -find lower and upper bounds on answer
            -figure out what ur trying to find and isolate it
            -see what observations you have and come up with more continuations
            -work backwards (in constructive, go from the goal to the start)
            -turn into prefix/suffix sum argument (often works if problem revolves around adjacent array elements)
            -instead of solving for answer, try solving for complement (ex, find n-(min) instead of max)
            -draw something
            -simulate a process
            -dont implement something unless if ur fairly confident its correct
            -after 3 bad submissions move on to next problem if applicable
            -do something instead of nothing and stay organized
            -write stuff down
         Random stuff to check when wa:
            -if code is way too long/cancer then reassess
            -switched N/M
            -int overflow
            -switched variables
            -wrong MOD
            -hardcoded edge case incorrectly
         Random stuff to check when tle:
            -continue instead of break
            -condition in for/while loop bad
         Random stuff to check when rte:
            -switched N/M
            -long to int/int overflow
            -division by 0
            -edge case for empty list/data structure/N=1
      */

	public static class Pair {
		int x;
		int y;

		public Pair(int x, int y) {
			this.x = x;
			this.y = y;
		}
	}

	public static void sort(int[] arr) {
		ArrayList<Integer> ls = new ArrayList<Integer>();
		for (int x : arr)
			ls.add(x);
		Collections.sort(ls);
		for (int i = 0; i < arr.length; i++)
			arr[i] = ls.get(i);
	}

	public static long gcd(long a, long b) {
		if (b == 0)
			return a;
		return gcd(b, a % b);
	}

	static boolean[] sieve(int N) {
		boolean[] sieve = new boolean[N + 1];
		for (int i = 2; i <= N; i++)
			sieve[i] = true;

		for (int i = 2; i <= N; i++) {
			if (sieve[i]) {
				for (int j = 2 * i; j <= N; j += i) {
					sieve[j] = false;
				}
			}
		}
		return sieve;
	}

	public static long power(long x, long y, long p) {
		long res = 1L;
		x = x % p;
		while (y > 0) {
			if ((y & 1) == 1)
				res = (res * x) % p;
			y >>= 1;
			x = (x * x) % p;
		}
		return res;
	}

	public static void print(int[] arr) {
		//for debugging only
		for (int x : arr)
			out.print(x + " ");
		out.println();
	}

	static class FastScanner {
		private int BS = 1 << 16;
		private char NC = (char) 0;
		private byte[] buf = new byte[BS];
		private int bId = 0, size = 0;
		private char c = NC;
		private double cnt = 1;
		private BufferedInputStream in;

		public FastScanner() {
			in = new BufferedInputStream(System.in, BS);
		}

		public FastScanner(String s) {
			try {
				in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
			} catch (Exception e) {
				in = new BufferedInputStream(System.in, BS);
			}
		}

		private char getChar() {
			while (bId == size) {
				try {
					size = in.read(buf);
				} catch (Exception e) {
					return NC;
				}
				if (size == -1)
					return NC;
				bId = 0;
			}
			return (char) buf[bId++];
		}

		public int nextInt() {
			return (int) nextLong();
		}

		public int[] readInts(int N) {
			int[] res = new int[N];
			for (int i = 0; i < N; i++) {
				res[i] = (int) nextLong();
			}
			return res;
		}

		public long[] readLongs(int N) {
			long[] res = new long[N];
			for (int i = 0; i < N; i++) {
				res[i] = nextLong();
			}
			return res;
		}

		public long nextLong() {
			cnt = 1;
			boolean neg = false;
			if (c == NC)
				c = getChar();
			for (; (c < '0' || c > '9'); c = getChar()) {
				if (c == '-')
					neg = true;
			}
			long res = 0;
			for (; c >= '0' && c <= '9'; c = getChar()) {
				res = (res << 3) + (res << 1) + c - '0';
				cnt *= 10;
			}
			return neg ? -res : res;
		}

		public double nextDouble() {
			double cur = nextLong();
			return c != '.' ? cur : cur + nextLong() / cnt;
		}

		public double[] readDoubles(int N) {
			double[] res = new double[N];
			for (int i = 0; i < N; i++) {
				res[i] = nextDouble();
			}
			return res;
		}

		public String next() {
			StringBuilder res = new StringBuilder();
			while (c <= 32)
				c = getChar();
			while (c > 32) {
				res.append(c);
				c = getChar();
			}
			return res.toString();
		}

		public String nextLine() {
			StringBuilder res = new StringBuilder();
			while (c <= 32)
				c = getChar();
			while (c != '\n') {
				res.append(c);
				c = getChar();
			}
			return res.toString();
		}

		public boolean hasNext() {
			if (c > 32)
				return true;
			while (true) {
				c = getChar();
				if (c == NC)
					return false;
				else if (c > 32)
					return true;
			}
		}
	}

	//	For Input.txt and Output.txt	
	//	FileInputStream in = new FileInputStream("input.txt");
	//	FileOutputStream out = new FileOutputStream("output.txt");
	//	PrintWriter pw = new PrintWriter(out);
	//	Scanner sc = new Scanner(in);
}