Problem: One day Nikita found the string containing letters "a" and "b" only. 

Nikita thinks that string is beautiful if it can be cut into 3 strings (possibly empty) without changing the order of the letters, where the 1-st and the 3-rd one contain only letters "a" and the 2-nd contains only letters "b".

Nikita wants to make the string beautiful by removing some (possibly none) of its characters, but without changing their order. What is the maximum length of the string he can get?

Input

The first line contains a non-empty string of length not greater than 5 000 containing only lowercase English letters "a" and "b". 

Output

Print a single integer — the maximum possible size of beautiful string Nikita can get.

Examples

Input

abba


Output

4

Input

bab


Output

2

Note

It the first sample the string is already beautiful.

In the second sample he needs to delete one of "b" to make it beautiful.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline int toInt(string s) {
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T>
inline string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}
template <class T>
inline T sqr(T x) {
  return x * x;
}
template <typename T>
T gcd(T x, T y) {
  if (y == 0)
    return x;
  else
    return gcd(y, x % y);
}
template <typename T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
template <typename T>
bool is_prime(T n) {
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return n != 1;
}
map<int, int> prime_factor(int n) {
  map<int, int> res;
  for (int i = 2; i * i <= n; i++) {
    while (n % i == 0) {
      ++res[i];
      n /= i;
    }
  }
  if (n != 1) res[n] = 1;
  return res;
}
int extgcd(int a, int b, int& x, int& y) {
  int d = a;
  if (b != 0) {
    d = extgcd(b, a % b, y, x);
    y -= (a / b) * x;
  } else {
    x = 1;
    y = 0;
  }
  return d;
}
long long mod_pow(long long x, long long n, long long mod) {
  if (n == 0) return 1;
  long long res = mod_pow(x * x % mod, n / 2, mod);
  if (n & 1) res = res * x % mod;
  return res;
}
vector<string> split(const string& str, char delim) {
  vector<string> res;
  size_t current = 0, found;
  while ((found = str.find_first_of(delim, current)) != string::npos) {
    res.push_back(string(str, current, found - current));
    current = found + 1;
  }
  res.push_back(string(str, current, str.size() - current));
  return res;
}
bool is_kadomatsu(int a, int b, int c) {
  if (a == b || a == c || b == c) return false;
  if (a > b && c > b) return true;
  if (a < b && c < b) return true;
  return false;
}
struct UF {
  int n;
  vector<int> d;
  UF() {}
  UF(int n) : n(n), d(n, -1) {}
  int root(int v) {
    if (d[v] < 0) return v;
    return d[v] = root(d[v]);
  }
  bool same(int a, int b) { return root(a) == root(b); }
  bool unite(int x, int y) {
    x = root(x);
    y = root(y);
    if (x == y) return false;
    if (size(x) < size(y)) swap(x, y);
    d[x] += d[y];
    d[y] = x;
    return true;
  }
  int size(int v) { return -d[root(v)]; }
};
vector<int> divisor(int n) {
  if (n == 1) return {};
  vector<int> res;
  for (int i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      res.emplace_back(i);
      if (i != 1 && i != n / i) res.emplace_back(n / i);
    }
  }
  return res;
}
struct Bellmanford {
  int n;
  struct edge {
    int from, to, cost;
  };
  vector<edge> E;
  vector<int> d;
  Bellmanford(int n) : n(n), d(n) { E.resize(n); }
  void add_edge(int x, int y, int cost) {
    edge e;
    e.from = x;
    e.to = y;
    e.cost = cost;
    E.push_back(e);
  }
  void shortest_path(int s) {
    for (int i = (0); i < (n); ++i) d[i] = (1000000000);
    d[s] = 0;
    while (true) {
      bool update = false;
      for (auto e : E) {
        if (d[e.from] != (1000000000) && d[e.to] > d[e.from] + e.cost) {
          d[e.to] = d[e.from] + e.cost;
          update = true;
        }
      }
      if (!update) break;
    }
  }
};
struct Dijkstra {
  int n;
  struct edge {
    int to;
    long long cost;
  };
  vector<vector<edge>> G;
  vector<long long int> d;
  Dijkstra(int n) : n(n), d(n) { G.resize(n); }
  void add_edge(int x, int y, long long cost) {
    edge e;
    e.to = y;
    e.cost = cost;
    G[x].push_back(e);
  }
  void shortest_path(int s) {
    for (int i = (0); i < (n); ++i) d[i] = 100000000000000000;
    d[s] = 0;
    priority_queue<pair<long long, int>, vector<pair<long long, int>>,
                   greater<pair<long long, int>>>
        que;
    que.push(make_pair(0, s));
    while (!que.empty()) {
      pair<int, int> p = que.top();
      que.pop();
      int v = p.second;
      if (d[v] < p.first) continue;
      for (auto e : G[v]) {
        if (d[e.to] > d[v] + e.cost) {
          d[e.to] = d[v] + e.cost;
          que.push(make_pair(d[e.to], e.to));
        }
      }
    }
  }
};
struct Segmenttree {
  int n;
  vector<pair<long long, int>> dat;
  Segmenttree() {}
  void init(long long n_) {
    n = 1;
    while (n < n_) n *= 2;
    dat.resize(2 * n - 1);
    for (int i = (0); i < (2 * n - 1); ++i)
      dat[i] = pair<long long, int>(-(1000000000), -(1000000000));
  }
  void update(int idx, long long val) {
    idx += n - 1;
    dat[idx] = make_pair(val, -(idx - n + 1));
    while (idx > 0) {
      idx = (idx - 1) / 2;
      dat[idx] = max(dat[idx * 2 + 1], dat[idx * 2 + 2]);
    }
  }
  pair<long long, int> query(int a, int b) { return query_seg(a, b, 0, 0, n); }
  pair<long long, int> query_seg(int a, int b, int k, int l, int r) {
    if (r <= a || b <= l)
      return pair<long long, int>(-(1000000000), -(1000000000));
    if (a <= l && r <= b)
      return dat[k];
    else {
      return max(query_seg(a, b, k * 2 + 1, l, (l + r) / 2),
                 query_seg(a, b, k * 2 + 2, (l + r) / 2, r));
    }
  }
};
int main() {
  string s;
  int A[5010], B[5010];
  memset(A, 0, sizeof(A));
  memset(B, 0, sizeof(B));
  cin >> s;
  int n = s.length();
  for (int i = (0); i < (n); ++i) {
    if (s[i] == 'a')
      A[i] = 1;
    else
      B[i] = 1;
  }
  for (int i = 1; i < n; i++) A[i] += A[i - 1];
  for (int i = 1; i < n; i++) B[i] += B[i - 1];
  int ans = A[n - 1];
  for (int R = (0); R < (n); ++R) {
    for (int L = (0); L < (R + 1); ++L) {
      int a = 0;
      if (L) a = A[L - 1];
      int c = A[n - 1] - A[R];
      int b = B[R];
      if (L) b -= B[L - 1];
      ans = max(ans, a + b + c);
    }
  }
  cout << (ans) << endl;
  ;
  return 0;
}