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.
# DEFINING SOME GOOD STUFF
from math import *
import threading
import sys
from collections import defaultdict
sys.setrecursionlimit(300000)
# threading.stack_size(10**5)  # remember it cause mle


mod = 10 ** 9
inf = 10 ** 15
yes = 'YES'
no = 'NO'

# ------------------------------FASTIO----------------------------
import os
import sys
from io import BytesIO, IOBase

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")



# _______________________________________________________________#

def npr(n, r):
    return factorial(n) // factorial(n - r) if n >= r else 0


def ncr(n, r):
    return factorial(n) // (factorial(r) * factorial(n - r)) if n >= r else 0


def lower_bound(li, num):
    answer = -1
    start = 0
    end = len(li) - 1

    while (start <= end):
        middle = (end + start) // 2
        if li[middle] >= num:
            answer = middle
            end = middle - 1
        else:
            start = middle + 1
    return answer  # min index where x is not less than num


def upper_bound(li, num):
    answer = -1
    start = 0
    end = len(li) - 1

    while (start <= end):
        middle = (end + start) // 2

        if li[middle] <= num:
            answer = middle
            start = middle + 1

        else:
            end = middle - 1
    return answer  # max index where x is not greater than num


def abs(x):
    return x if x >= 0 else -x


def binary_search(li, val):
    # print(lb, ub, li)
    ans = -1
    lb = 0
    ub = len(li) - 1
    while (lb <= ub):
        mid = (lb + ub) // 2
        # print('mid is',mid, li[mid])
        if li[mid] > val:
            ub = mid - 1
        elif val > li[mid]:
            lb = mid + 1
        else:
            ans = mid  # return index
            break
    return ans


def kadane(x):  # maximum sum contiguous subarray
    sum_so_far = 0
    current_sum = 0
    for i in x:
        current_sum += i
        if current_sum < 0:
            current_sum = 0
        else:
            sum_so_far = max(sum_so_far, current_sum)
    return sum_so_far


def pref(li):
    pref_sum = [0]
    for i in li:
        pref_sum.append(pref_sum[-1] + i)
    return pref_sum


def SieveOfEratosthenes(n):
    prime = [True for i in range(n + 1)]
    p = 2
    li = []
    while (p * p <= n):
        if (prime[p] == True):
            for i in range(p * p, n + 1, p):
                prime[i] = False
        p += 1

    for p in range(2, len(prime)):
        if prime[p]:
            li.append(p)
    return li


def primefactors(n):
    factors = []
    while (n % 2 == 0):
        factors.append(2)
        n //= 2
    for i in range(3, int(sqrt(n)) + 1, 2):  # only odd factors left
        while n % i == 0:
            factors.append(i)
            n //= i
    if n > 2:  # incase of prime
        factors.append(n)
    return factors


def prod(li):
    ans = 1
    for i in li:
        ans *= i
    return ans


def dist(a, b):
    d = abs(a[1] - b[1]) + abs(a[2] - b[2])
    return d


def power_of_n(x, n):
    cnt = 0
    while (x % n == 0):
        cnt += 1
        x //= n
    return cnt

def check(u,r,d,l):
    f = 1
    if u == 0 or d == 0:
        if l == n or r == n:
            f = 0
            # print('1')
    if l == 0 or r == 0:
        if d == n or u == n:
            f = 0
            # print('2')
    if u == 0 and d == 0:
        if l >= n-1 or r >= n-1:
            f = 0
            # print('3')
    if l == 0 and r == 0:
        if d >= n-1 or u >= n-1:
            f = 0
            # print(4)
    return f




# _______________________________________________________________#


# def main():

for _ in range(1):
# for _ in range(int(input()) if True else 1):
    n = int(input())
#     n, u, r, d, l = map(int, input().split())
    a = list(map(int, input().split()))
    # b = list(map(int, input().split()))
    # c = list(map(int, input().split()))
    # s = list(input())
    # s = input()
    # t = input()
    # c = list(map(int,input().split()))
    # adj = graph(n,m)
    dp = [[0, 0, 0]]*(n+1)
    for i in range(1, n+1):
        dp[i] = [min(dp[i-1]) + 1] + dp[i][1:]
        # print(dp)
        if a[i-1] == 0:
            dp[i][1] = dp[i-1][1] + 1
            dp[i][2] = dp[i-1][2] + 1
        elif a[i-1] == 1:
            dp[i][1] = min(dp[i - 1][0], dp[i-1][2])
            dp[i][2] = dp[i - 1][2] + 1
        elif a[i-1] == 2:
            dp[i][1] = dp[i-1][1]+1
            dp[i][2] = min(dp[i-1][0], dp[i-1][1])
        else:
            dp[i][1] = min(dp[i-1][0], dp[i-1][2])
            dp[i][2] = min(dp[i-1][0], dp[i-1][1])
    # print(dp)
    print(min(dp[-1]))









    
            





























'''		
t = threading.Thread(target=main)
t.start()
t.join()
'''
