#include<iostream>
#include<cstring>
#include<vector>
using namespace std;
using LL = long long;
const int M = 40;
int dp[M + 1][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[M + 1];
    fib[0] = fib[1] = 1;
    for(int i = 2; i <= M; 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 <= M; 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[i][j:]位置匹配t[k:]能匹配的长度和需要的代价
    auto dfs = [&](auto &&dfs, int i, int j, int k) -> pair<int, int> {
        if (k == n) return {0, 0};
        if (i == 0){
            if (t[k] == 'b') return {1, 1};
            return {0, 1};
        }
        if (i == 1){
            if (t[k] == 'a') return {1, 1};
            return {0, 1};
        }

        if (j == 0){
            if (k + dp[i][k] < n) return {dp[i][k], fib[i]};
            if (k + dp[i - 1][k] < n){
                auto [a, b] = dfs(dfs, i - 2, 0, k + dp[i - 1][k]);
                return {a + dp[i - 1][k], b + fib[i - 1]};
            }
            return dfs(dfs, i - 1, 0, k);
        }
        if (j < fib[i - 1]){
            auto [a, b] = dfs(dfs, i - 1, j, k);
            if (k + a == n) return {a, b};
            auto [c, d] = dfs(dfs, i - 2, 0, k + a);
            return {a + c, b + d};
        }
        return dfs(dfs, i - 2, j - fib[i - 1], k);
    };

    const int N = 500000;

    int ans = 1e9;
    for(int i = 0; i < fib[28]; i++){
        ans = min(ans, dfs(dfs, 30, i, 0).second);
    }
    cout << ans << '\n';

}