Problem:
Score : 700 points

Problem Statement

N cells are arranged in a row.
Some of them may contain tokens.
You are given a string s that consists of 0s and 1s.
If the i-th character of s is 1, the i-th cell (from left) contains a token.
Otherwise, it doesn't contain a token.

Snuke wants to perform the following operation as many times as possible.
In each operation, he chooses three consecutive cells.
Let's call the cells X, Y, Z from left to right.
In order for the operation to be valid, both X and Z must contain tokens and Y must not contain a token.
Then, he removes these two tokens and puts a new token on Y.

How many operations can he perform if he performs operations in the optimal way?

Constraints

1 \leq N \leq 500,000

|s| = N

Each character in s is either 0 or 1.

Input

Input is given from Standard Input in the following format:

N
s

Output

Print the answer.

Sample Input 1

7
1010101

Sample Output 1

2

For example, he can perform two operations in the following way:

Perform an operation on the last three cells. Now the string that represents tokens becomes 1010010.

Perform an operation on the first three cells. Now the string that represents tokens becomes 0100010.

Note that the choice of operations matters.
For example, if he chooses three cells in the middle first, he can perform no more operations.

Sample Input 2

50
10101000010011011110001001111110000101010111100110

Sample Output 2

10

Solution in PyPy3 (7.3.0) (judge verdict: Runtime Error)

from collections import defaultdict, deque, Counter
from heapq import heappush, heappop, heapify
import math
import bisect
import random
from itertools import permutations, accumulate, combinations, product
import sys
import string
from bisect import bisect_left, bisect_right
from math import factorial, ceil, floor
from operator import mul
from functools import reduce


sys.setrecursionlimit(2147483647)
INF = 10 ** 13
def LI(): return list(map(int, sys.stdin.readline().split()))
def I(): return int(sys.stdin.readline())
def LS(): return sys.stdin.readline().rstrip().split()
def S(): return sys.stdin.readline().rstrip()
def IR(n): return [I() for i in range(n)]
def LIR(n): return [LI() for i in range(n)]
def SR(n): return [S() for i in range(n)]
def LSR(n): return [LS() for i in range(n)]
def SRL(n): return [list(S()) for i in range(n)]
def MSRL(n): return [[int(j) for j in list(S())] for i in range(n)]
mod = 10 ** 9 + 7


n = I()
s = list(S())
ans = 0
one_cnt = 0
z_cnt = 0
dp = []
flg = 0
for i in range(n):
    if s[i] == "1":
        one_cnt += 1
        z_cnt = 0
        if flg:
            ans += max(dp[-1])
            dp = []
            flg = 0

        if i == n - 1 or s[i + 1] == "0":
            # 手付かず、左端だけ消した、右端だけ残す、全て消した。
            if not dp:
                dp += [[0, -INF, 0, -INF]]
            else:
                if one_cnt != 1:
                    dp += [[max(dp[-1]), max(dp[-1][0] + pre, dp[-1][2] + 1),
                            max(dp[-1][2] + one_cnt - 1, pre + dp[-1][0]), dp[-1][2] + one_cnt]]
                else:
                    dp += [[max(dp[-1]), max(dp[-1][2] + 1, dp[-1][0] + pre),
                            max(dp[-1]), max(dp[-1][2] + 1, dp[-1][0] + pre)]]
            pre = one_cnt
    else:
        z_cnt += 1
        one_cnt = 0
        if z_cnt >= 2:
            flg = 1

if dp:
    ans += max(dp[-1])

if ans == -INF:
    print(0)
else:
    print(ans)