#include <bits/stdc++.h>
using namespace std;
const int P = 1e9 + 7, INF = 0x3f3f3f3f;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long qpow(long long a, long long n) {
  long long r = 1 % P;
  for (a %= P; n; a = a * a % P, n >>= 1)
    if (n & 1) r = r * a % P;
  return r;
}
long long inv(long long first) {
  return first <= 1 ? 1 : inv(P % first) * (P - P / first) % P;
}
inline int rd() {
  int first = 0;
  char p = getchar();
  while (p < '0' || p > '9') p = getchar();
  while (p >= '0' && p <= '9') first = first * 10 + p - '0', p = getchar();
  return first;
}
const int N = 1e6 + 50;
int n, mi[N], sum[N];
char s[N];
int solve(int first) {
  int tot = 0, now = 0, mx = 0;
  for (int i = 1; i <= n; ++i) {
    if (s[i] == '?') {
      if (mi[i] - tot - 2 >= first)
        --now, tot += 2;
      else
        ++now;
    } else if (s[i] == '0')
      --now;
    else
      ++now;
    mx = max(mx, now);
  }
  return mx;
}
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  int mx = 0;
  for (int i = 1; i <= n; ++i) sum[i] = sum[i - 1] + (s[i] == '0' ? -1 : 1);
  mi[n] = sum[n];
  for (int i = n - 1; i >= 1; --i) mi[i] = min(mi[i + 1], sum[i]);
  printf("%d\n", min(solve(mi[1]) - mi[1], solve(mi[1] - 1) - mi[1] + 1));
}
