// This function is working in O(max(s1.length(), s2.length())
int recur(string s1, string s2, int x, int y)
{
    if (x >= s1.length() || y >= s2.length())
        return 0;
    int t1 = 0, t2 = 0;

    if (s1[x] == s2[y])
        t1 = 1 + recur(s1, s2, x + 1, y + 1);
    else
        t2 = recur(s1, s2, x, y + 1);

    return max(t1, t2);
}

// storing the power to 2 from 0-40
vector<string> make_power_of_2()
{
    vector<string> vs;
    vs.pb("1");
    int ans = 1;
    for (int i = 1; i <= 40; i++)
    {
        ans = ans * 2;
        vs.pb(to_string(ans));
    }
    return vs;
}

void solve()
{
    vector<string> vs = make_power_of_2();
    int n;
    cin >> n;

    string str = to_string(n);
    int ans = INT_MAX;
    int flag = 0;
    for (int i = 0; i < vs.size(); i++)
    {
        int z = recur(vs[i], str, 0, 0);
        ans = min(ans, (vs[i].length() + str.length() - 2 * z));
    }
    cout << min(ans, (int)(str.length() + 1)) << endl;
}