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 java.util.*;

public class Main {
	static int mod = 998244353;
  static int size = 200000;
	static long[] fac = new long[size];
	static long[] finv = new long[size];
	static long[] inv = new long[size];
	static int INF = Integer.MAX_VALUE;

 	public static void main(String[] args){
		Scanner scanner = new Scanner(System.in);
		int n = scanner.nextInt();
		int[] a = new int[n+1];
		for(int i = 1; i <= n; i++){
			a[i] = scanner.nextInt();
		}
		//dp[i][j]:= i日目にj(j=0 休む、j=1 ジム、j=2 コンテスト)をする時の
		//休みの回数の最小値
		int[][] dp = new int[n+1][3];
		for(int i = 1; i <= n; i++){
			for(int j = 0; j < 3; j++){
				dp[i][j] = INF;
			}
		}
		dp[1][0] = 1;
		for(int i = 1; i <= n; i++){
			if(a[i] == 0){
				dp[i][0] = Math.min(Math.min(dp[i-1][0],dp[i-1][1]),dp[i-1][2])+1;
			}
			if(a[i] == 1){
				dp[i][0] = Math.min(Math.min(dp[i-1][0],dp[i-1][1]),dp[i-1][2])+1;
				dp[i][2] = Math.min(dp[i-1][0],dp[i-1][1]);
			}
			if(a[i] == 2){
				dp[i][0] = Math.min(Math.min(dp[i-1][0],dp[i-1][1]),dp[i-1][2])+1;
				dp[i][1] = Math.min(dp[i-1][0],dp[i-1][2]);
			}
			if(a[i] == 3){
				dp[i][0] = Math.min(Math.min(dp[i-1][0],dp[i-1][1]),dp[i-1][2])+1;
				dp[i][1] = Math.min(dp[i-1][0],dp[i-1][2]);
				dp[i][2] = Math.min(dp[i-1][0],dp[i-1][1]);
			}
		}
		int min = 1000;
		for(int i = 0; i < 3; i++){
			min = Math.min(min,dp[n][i]);
		}
		System.out.println(min);
	}
	public static class Edge{
		int to;
		Edge(int to){
			this.to = to;
		}
	}
	public static void swap(long a, long b){
		long tmp = 0;
		if(a > b){
			tmp = a;
			a = b;
			b = tmp;
		}
	}
  static class Pair implements Comparable<Pair>{
    int first, second;
    Pair(int a, int b){
        first = a;
        second = b;
    }
    @Override
    public boolean equals(Object o){
        if (this == o) return true;
        if (!(o instanceof Pair)) return false;
        Pair p = (Pair) o;
        return first == p.first && second == p.second;
    }
    @Override
    public int compareTo(Pair p){
        //return first == p.first ? second - p.second : first - p.first; //firstで昇順にソート
        //return (first == p.first ? second - p.second : first - p.first) * -1; //firstで降順にソート
        return second == p.second ? first - p.first : second - p.second;//secondで昇順にソート
        //return (second == p.second ? first - p.first : second - p.second)*-1;//secondで降順にソート
    }
  }

  //繰り返し二乗法
  public static long pow(long x, long n){
    long ans = 1;
    while(n > 0){
      if((n & 1) == 1){
        ans = ans * x;
        ans %= mod;
      }
      x = x * x % mod;
      n >>= 1;
    }
    return ans;
  }

	public static long div(long x, long y){
		return (x*pow(y, mod-2))%mod;
	}

  //fac, inv, finvテーブルの初期化、これ使う場合はinitComb()で初期化必要
	public static  void initComb(){
		fac[0] = finv[0] = inv[0] = fac[1] = finv[1] = inv[1] = 1;
		for (int i = 2; i < size; ++i) {
			fac[i] = fac[i - 1] * i % mod;
			inv[i] = mod - (mod / i) * inv[(int) (mod % i)] % mod;
			finv[i] = finv[i - 1] * inv[i] % mod;
		}
	}

	//nCk % mod
	public static long comb(int n, int k){
		return fac[n] * finv[k] % mod * finv[n - k] % mod;
	}

	//n! % mod
	public static long fact(int n){
		return fac[n];
	}

	//(n!)^-1 with % mod
	public static long finv(int n){
		return finv[n];
	}

  static class UnionFind {
    int[] parent;
    public UnionFind(int size) {
      parent = new int[size];
      Arrays.fill(parent, -1);
    }
    public boolean unite(int x, int y) {
      x = root(x);
      y = root(y);
      if (x != y) {
        if (parent[y] < parent[x]) {
          int tmp = y;
          y = x;
          x = tmp;
        }
        parent[x] += parent[y];
        parent[y] = x;
        return true;
      }
      return false;
    }
    public boolean same(int x, int y) {
      return root(x) == root(y);
    }
    public int root(int x) {
      return parent[x] < 0 ? x : (parent[x] = root(parent[x]));
    }
    public int size(int x) {
      return -parent[root(x)];
    }
  }
	public static int upperBound(long[] array, long value) {
			 int low = 0;
			 int high = array.length;
			 int mid;
			 while( low < high ) {
					 mid = ((high - low) >>> 1) + low; // (high + low) / 2
					 if( array[mid] <= value ) {
							 low = mid + 1;
					 } else {
							 high = mid;
					 }
			 }
			 return low;
	 }
	 public static final int lowerBound(final long[] arr, final long value) {
    int low = 0;
    int high = arr.length;
    int mid;
    while (low < high) {
        mid = ((high - low) >>> 1) + low;    //(low + high) / 2 (オーバーフロー対策)
        if (arr[mid] < value) {
            low = mid + 1;
        } else {
            high = mid;
        }
    }
    return low;
}
  //n,mの最大公約数
  public static long gcd(long n, long m){
    if(m > n) return gcd(m,n);
    if(m == 0) return n;
    return gcd(m, n%m);
  }
	//3要素のソート
	private class Pair2 implements Comparable<Pair2> {
		String s;
		int p;
		int index;
		public Pair2(String s, int p, int index) {
				this.s = s;
				this.p = p;
				this.index = index;
		}

		public int compareTo(Pair2 other) {
				if (s.equals(other.s)) {
						return other.p - this.p;
				}
				return this.s.compareTo(other.s);
		}
	}
	//c -> intに変換
public static long c2i(char c){
	if('A' <= c && c <= 'Z'){
		return c - 'A';
	}else{
		return c - 'a' + 26;
	}
}
// int -> toに変換
public static char i2c(long i){
	if(0 <= i && i < 26){
			return (char)(i + 'A');
	}else{
			return (char)(i + 'a' - 26);
	}
}

}
