Problem:
Score : 800 points

Problem Statement

Given is a string S, where each character is 0, 1, or ?.

Consider making a string S' by replacing each occurrence of ? with 0 or 1 (we can choose the character for each ? independently).
Let us define the unbalancedness of S' as follows:

(The unbalancedness of S') = \max \{ The absolute difference between the number of occurrences of 0 and 1 between the l-th and r-th character of S (inclusive) :\ 1 \leq l \leq r \leq |S|\}

Find the minimum possible unbalancedness of S'.

Constraints

1 \leq |S| \leq 10^6

Each character of S is 0, 1, or ?.

Input

Input is given from Standard Input in the following format:

S

Output

Print the minimum possible unbalancedness of S'.

Sample Input 1

0??

Sample Output 1

1

We can make S'= 010 and achieve the minimum possible unbalancedness of 1.

Sample Input 2

0??0

Sample Output 2

2

Sample Input 3

??00????0??0????0?0??00??1???11?1?1???1?11?111???1

Sample Output 3

4

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define FOR(i, begin, end) for(int i=(begin);i<(end);i++)
#define REP(i, n) FOR(i,0,n)
#define IFOR(i, begin, end) for(int i=(end)-1;i>=(begin);i--)
#define IREP(i, n) IFOR(i,0,n)
#define Sort(v) sort(v.begin(), v.end())
#define Reverse(v) reverse(v.begin(), v.end())
#define all(v) v.begin(),v.end()
#define SZ(v) ((int)v.size())
#define Lower_bound(v, x) distance(v.begin(), lower_bound(v.begin(), v.end(), x))
#define Upper_bound(v, x) distance(v.begin(), upper_bound(v.begin(), v.end(), x))
#define Max(a, b) a = max(a, b)
#define Min(a, b) a = min(a, b)
#define bit(n) (1LL<<(n))
#define debug(x) cout << #x << "=" << x << endl;
#define vdebug(v) { cout << #v << "=" << endl; REP(i_debug, v.size()){ cout << v[i_debug] << ","; } cout << endl; }
#define mdebug(m) { cout << #m << "=" << endl; REP(i_debug, m.size()){ REP(j_debug, m[i_debug].size()){ cout << m[i_debug][j_debug] << ","; } cout << endl;} }
#define Return(ans) { cout << (ans) << endl; return 0; }
#define pb push_back
#define fi first
#define se second
#define int long long
#define INF 1000000000000000000
template<typename T> istream &operator>>(istream &is, vector<T> &v){ for (auto &x : v) is >> x; return is; }
template<typename T> ostream &operator<<(ostream &os, vector<T> &v){ for(int i = 0; i < v.size(); i++) { cout << v[i]; if(i != v.size() - 1) cout << endl; }; return os; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, pair<T1, T2> p){ cout << '(' << p.first << ',' << p.second << ')'; return os; }
template<typename T> void Out(T x) { cout << x << endl; }
template<typename T1, typename T2> void Ans(bool f, T1 y, T2 n) { if(f) Out(y); else Out(n); }

using vec = vector<int>;
using mat = vector<vec>;
using Pii = pair<int, int>;
using v_bool = vector<bool>;
using v_Pii = vector<Pii>;

//int dx[4] = {1,0,-1,0};
//int dy[4] = {0,1,0,-1};
//char d[4] = {'D','R','U','L'};

const int mod = 1000000007;
//const int mod = 998244353;



signed main(){

    string S; cin >> S;
    int N = SZ(S);

    REP(q, 2){
        bool ok = true;
        REP(i, N){
            if(S[i] == '0' + (i % 2 + q) % 2) ok = false;
        }
        if(ok){
            Out(1);
            return 0;
        }
    }

    int n0 = 1, n1 = N;
    while(n1 - n0 > 1){
        int n = (n0 + n1) / 2;
        bool ok = true;

        REP(q, 2){
            int c = 0;
            int m = 0, M = 0, now = 0;
            REP(i, N){
                if(S[i] == '?') c++;
                else{
                    if(c > 0){
                        if(c > n){
                            if((c - n) % 2 == 0) c = n;
                            else c = n - 1;
                        }
                        now -= c;
                        c = 0;

                        Max(M, now - m);
                        Min(m, now);
                    }

                    if(S[i] == '0' + q) now++;
                    else now--;
                    Max(M, now - m);
                    Min(m, now);
                }
            }
            if(c > 0){
                if(c > n){
                    if((c - n) % 2 == 0) c = n;
                    else c = n - 1;
                }
                now -= c;
                c = 0;

                Max(M, now - m);
                Min(m, now);
            }

            if(M > n) ok = false;
        }

        if(ok) n1 = n;
        else n0 = n;
    }
    Out(n1);

    return 0;
}