N cells are arranged in a row. Some of them may contain tokens. You are given a string s that consists of `0`s and `1`s. 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.

Examples

Input

7
1010101


Output

2


Input

50
10101000010011011110001001111110000101010111100110


Output

10
#include <bits/stdc++.h>
using namespace std;
using lint = long long int;
template<class T = int> using V = vector<T>;
template<class T = int> using VV = V< V<T> >;
template<class T> void assign(V<T>& v, int n, const T& a = T()) { v.assign(n, a); }
template<class T, class... U> void assign(V<T>& v, int n, const U&... u) { v.resize(n); for (auto&& i : v) assign(i, u...); }

int main() {
  cin.tie(NULL); ios::sync_with_stdio(false);
  int n; cin >> n;
  string s; cin >> s;
  s = "00" + s, n += 2;
  V<> iz(n);
  int t = 0;
  for (int i = 0; i < n; i++) if (s[i] == '0') t = 0;
  else iz[i] = ++t;
  V<> dp(n + 1);
  for (int i = 2; i < n; i++) {
    dp[i + 1] = dp[i];
    if (s.substr(i - 2, 3) == "101") dp[i + 1] = max({dp[i + 1], dp[i - iz[i - 2] - 1] + iz[i - 2], dp[i - iz[i - 2]] + iz[i - 2] - 1});
    if (s.substr(i - 1, 2) == "11" and s.substr(i - iz[i] - 1, 2) == "10") dp[i + 1] = max(dp[i + 1], dp[i - iz[i] - 1] + iz[i]);
  }
  cout << dp[n] << '\n';
}