Problem: Given is a string S consisting of `0` and `1`. Find the number of strings, modulo 998244353, that can result from applying the following operation on S zero or more times:

* Remove the two characters at the beginning of S, erase one of them, and reinsert the other somewhere in S. This operation can be applied only when S has two or more characters.

Constraints

* 1 \leq |S| \leq 300
* S consists of `0` and `1`.

Input

Input is given from Standard Input in the following format:


S


Output

Print the number of strings, modulo 998244353, that can result from applying the operation on S zero or more times.

Examples

Input

0001


Output

8


Input

110001


Output

24


Input

11101111011111000000000110000001111100011111000000001111111110000000111111111


Output

697354558

Solution (in CPP):

Solution:
/*
    author:  Maksim1744
    created: 20.06.2020 16:18:52
*/

#include <bits/stdc++.h>

using namespace std;

#define ll   long long
#define ld   long double

#define mp   make_pair
#define pb   push_back
#define eb   emplace_back

#define sum(a)     ( accumulate ((a).begin(), (a).end(), 0ll))
#define mine(a)    (*min_element((a).begin(), (a).end()))
#define maxe(a)    (*max_element((a).begin(), (a).end()))
#define mini(a)    ( min_element((a).begin(), (a).end()) - (a).begin())
#define maxi(a)    ( max_element((a).begin(), (a).end()) - (a).begin())
#define lowb(a, x) ( lower_bound((a).begin(), (a).end(), (x)) - (a).begin())
#define uppb(a, x) ( upper_bound((a).begin(), (a).end(), (x)) - (a).begin())

template<typename T>             vector<T>& operator--            (vector<T>& v){for (auto& i : v) --i;            return  v;}
template<typename T>             vector<T>& operator++            (vector<T>& v){for (auto& i : v) ++i;            return  v;}
template<typename T>             istream& operator>>(istream& is,  vector<T>& v){for (auto& i : v) is >> i;        return is;}
template<typename T>             ostream& operator<<(ostream& os,  vector<T>& v){for (auto& i : v) os << i << ' '; return os;}
template<typename T, typename U> istream& operator>>(istream& is, pair<T, U>& p){is >> p.first >> p.second;        return is;}
template<typename T, typename U> ostream& operator<<(ostream& os, pair<T, U>& p){os << p.first << ' ' << p.second; return os;}
template<typename T, typename U> pair<T,U> operator-(pair<T,U> a, pair<T,U> b){return mp(a.first-b.first, a.second-b.second);}
template<typename T, typename U> pair<T,U> operator+(pair<T,U> a, pair<T,U> b){return mp(a.first+b.first, a.second+b.second);}
template<typename T, typename U> void umin(T& a, U b){if (a > b) a = b;}
template<typename T, typename U> void umax(T& a, U b){if (a < b) a = b;}

#ifdef HOME
#define TAG_LENGTH       25
#define LR_LEFT          left
#define LR_RIGHT         right
#define LR_VALUE         value
#define LR_SECOND_VALUE  mn
#include "C:/C++ libs/print.cpp"
#else
#define showl            42;
#define shows            42;
#define show(...)        42;
#define showm(...)       42;
#define showt(...)       42;
#define printTree(...)   42;
#define printGraph(...)  42;
#define printLRTree(...) 42;
#define printMatrix(...) 42;
#define printWGraph(...) 42;
#define debug     if (false)
#endif

template<int P>
struct Modular {
    using value_type = decltype(P);
    value_type value;

    Modular(ll k = 0) : value(norm(k)) {}

    Modular& operator += (const Modular& m)       { value += m.value; if (value >= P) value -= P; return *this; }
    Modular  operator +  (const Modular& m) const { Modular r = *this; return r += m; }

    Modular& operator -= (const Modular& m)       { value -= m.value; if (value <  0) value += P; return *this; }
    Modular  operator -  (const Modular& m) const { Modular r = *this; return r -= m; }
    Modular  operator -                     () const { return Modular(-value); }

    Modular& operator *= (const Modular &m)       { value = value * 1ll * m.value % P; return *this; }
    Modular  operator *  (const Modular& m) const { Modular r = *this; return r *= m; }

    Modular& operator /= (const Modular &m)       { return *this *= m.inv(); }
    Modular  operator /  (const Modular& m) const { Modular r = *this; return r /= m; }

    Modular& operator ++                    ()       { return *this += 1; }
    Modular& operator --                    ()       { return *this -= 1; }
    Modular  operator ++                 (int)       { Modular r = *this; r += 1; return r; }
    Modular  operator --                 (int)       { Modular r = *this; r -= 1; return r; }

    bool        operator == (const Modular& m) const { return value == m.value; }
    bool        operator != (const Modular& m) const { return value != m.value; }

    value_type norm(ll k) {
        if (!(-P <= k && k < P)) k %= P;
        if (k < 0) k += P;
        return k;
    }

    Modular inv() const {
        value_type a = value, b = P, x = 0, y = 1;
        while (a != 0) { value_type k = b / a; b -= k * a; x -= k * y; swap(a, b); swap(x, y); }
        return Modular<P>(x);
    }
};

template<int P> Modular<P> pow(Modular<P> m, ll p) {
    Modular<P> r(1);
    while (p) {
        if (p & 1) r *= m;
        m *= m;
        p >>= 1;
    }
    return r;
}

template<int P> ostream& operator << (ostream& o, const Modular<P> m) { return o << m.value; }
template<int P> istream& operator >> (istream& i,       Modular<P> m) { ll k; i >> k; m.value = m.norm(k); return i; }
template<int P> string   to_string(const Modular<P>& m) { return to_string(m.value); }

// using Mint = Modular<1000000007>;
using Mint = Modular<998244353>;
// using Mint = long double;

int main() {
    ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);

    string s;
    cin >> s;
    int N = s.size() + 3;
    vector<vector<Mint>> C(N * 2 + 5, vector<Mint>(N * 2 + 5, 0));
    for (int i = 0; i < C.size(); ++i) {
        C[0][i] = 0;
        C[i][0] = 1;
    }
    for (int i = 1; i < C.size(); ++i) {
        for (int j = 1; j < C.size(); ++j) {
            C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
        }
    }
    vector<vector<vector<Mint>>> dp(N, vector<vector<Mint>>(N, vector<Mint>(N, 0)));
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < N; ++j) {
            dp[s.size()][i][j] = C[i + j][i];
        }
    }
    for (int i = s.size() - 1; i >= 0; --i) {
        for (int j = 0; j < N; ++j) {
            for (int k = 0; k < N; ++k) {
                if (s[i] == '1') {
                    dp[i][j][k] = (j == 0 ? Mint(0) : dp[i][j - 1][k]) + dp[i + 1][j][k];
                } else {
                    dp[i][j][k] = dp[i + 1][j][k] + (k == 0 ? Mint(0) : dp[i][j][k - 1]);
                }
            }
        }
    }
    vector<vector<vector<bool>>> can(N, vector<vector<bool>>(N, vector<bool>(N, 0)));
    can[0][0][0] = true;
    for (int i = 1; i <= s.size(); ++i) {
        for (int j = 0; j < N; ++j) {
            for (int k = 0; k < N; ++k) {
                if (i >= 2) {
                    if (s[i - 1] == '0' || s[i - 2] == '0')
                        if (j >= 1)
                            can[i][j][k] = can[i][j][k] || can[i - 2][j - 1][k];
                    if (s[i - 1] == '1' || s[i - 2] == '1')
                        if (k >= 1)
                            can[i][j][k] = can[i][j][k] || can[i - 2][j][k - 1];
                }
                if (j > 0 || k > 0)
                    can[i][j][k] = can[i][j][k] || can[i - 1][j][k];

                if (s[i - 1] == '0' && j > 0 && k + 1 < N) {
                    can[i][j][k] = can[i][j][k] || can[i - 1][j - 1][k + 1];
                }
                if (s[i - 1] == '1' && k > 0 && j + 1 < N) {
                    can[i][j][k] = can[i][j][k] || can[i - 1][j + 1][k - 1];
                }
            }
        }
    }
    function<void(int, int)> dfs = [&](int i, int j) {
        can[s.size()][i][j] = true;
        if (i > 0 && !can[s.size()][i - 1][j])
            dfs(i - 1, j);
        if (j > 0 && !can[s.size()][i][j - 1])
            dfs(i, j - 1);
    };
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < N; ++j) {
            if (can[s.size()][i][j])
                dfs(i, j);
        }
    }
    Mint ans = 0;
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < N; ++j) {
            if (i == 0 && j == 0) continue;
            int ind = -1;
            int i0 = 0, j0 = 0;
            for (int x = s.size(); x >= 0; --x) {
                if (i < i0 || j < j0) break;
                if (can[x][i - i0][j - j0]) {
                    ind = x;
                    break;
                }
                if (x != 0) {
                    i0 += (s[x - 1] == '0');
                    j0 += (s[x - 1] == '1');
                }
            }
            if (ind != -1) {
                // show(i, j);
                // show(i0, j0);
                // show(ind, dp[ind][i-i0][j-j0]);
                // shows;
                ans += dp[ind][i - i0][j - j0];
            }
        }
    }
    cout << ans << endl;

    return 0;
}