#include<iostream>
#include<cstring>
#include<vector>
using namespace std;
using LL = long long;
int dp[31][150005];

int main(){

#ifdef LOCAL
    freopen("data.in", "r", stdin);
    freopen("data.out", "w", stdout);
#endif

    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(0);

    int fib[31];
    fib[0] = fib[1] = 1;
    for(int i = 2; i <= 30; i++){
        fib[i] = fib[i - 1] + fib[i - 2];
    }

    string t;
    cin >> t;
    const int n = t.size();
    for(int i = 0; i < n; i++){
        dp[0][i] = (t[i] == 'b');
        dp[1][i] = (t[i] == 'a');
    }
    for(int i = 2; i <= 30; i++){
        for(int j = 0; j < n; j++){
            dp[i][j] = dp[i - 1][j] + dp[i - 2][j + dp[i - 1][j]];
        }
    }

    // 在s[k][x:]位置匹配t[y:]能匹配的长度和需要的代价
    auto dfs = [&](auto &&dfs, int u, int x, int y) -> pair<int, int> {
        if (y == n) 
            return {0, 0};
        if (u == 0){
            if (t[y] == 'b') return {1, 1};
            return {0, 1};
        }
        if (u == 1){
            if (t[y] == 'a') return {1, 1};
            return {0, 1};
        }
        if (x == 0){
            if (y + dp[u][y] < n) return {dp[u][y], fib[u]};
            if (y + dp[u - 1][y] < n){
                auto [a, b] = dfs(dfs, u - 2, 0, y + dp[u - 1][y]);
                return {a + dp[u - 1][y], b + fib[u - 1]};
            }
            return dfs(dfs, u - 1, 0, y);
        }
        if (x < fib[u - 1]){
            auto [a, b] = dfs(dfs, u - 1, x, y);
            if (y + a == n) return {a, b};
            auto [c, d] = dfs(dfs, u - 2, 0, y + a);
            return {a + c, b + d};
        }
        return dfs(dfs, u - 2, x - fib[u - 1], y);
    };

    const int N = 500000;

    int ans = 1e9;
    for(int i = 0; i < N; i++){
        ans = min(ans, dfs(dfs, 30, i, 0).second);
    }
    cout << ans << '\n';

}