#include <bits/stdc++.h>
using namespace std;
//#pragma GCC optimize("Ofast")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
 
#define ms(s, n) memset(s, n, sizeof(s))
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define FORd(i, a, b) for (int i = (a) - 1; i >= (b); --i)
#define FORall(it, a) for (__typeof((a).begin()) it = (a).begin(); it != (a).end(); it++)
#define sz(a) int((a).size())
#define present(t, x) (t.find(x) != t.end())
#define all(a) (a).begin(), (a).end()
#define uni(a) (a).erase(unique(all(a)), (a).end())
#define pb push_back
#define pf push_front
#define mp make_pair
#define fi first
#define se second
#define prec(n) fixed<<setprecision(n)
#define bit(n, i) (((n) >> (i)) & 1)
#define bitcount(n) __builtin_popcountll(n)
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pi;
typedef vector<int> vi;
typedef vector<pi> vii;
//const int MOD = (int) 1e9 + 7;
const int MOD = 119 << 23 | 1;
const int FFTMOD = 119 << 23 | 1;
const int INF = (int) 1e9 + 23111992;
const ll LINF = (ll) 1e18 + 23111992;
const ld PI = acos((ld) -1);
const ld EPS = 1e-9;
inline ll gcd(ll a, ll b) {ll r; while (b) {r = a % b; a = b; b = r;} return a;}
inline ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
inline ll fpow(ll n, ll k, int p = MOD) {ll r = 1; for (; k; k >>= 1) {if (k & 1) r = r * n % p; n = n * n % p;} return r;}
template<class T> inline int chkmin(T& a, const T& val) {return val < a ? a = val, 1 : 0;}
template<class T> inline int chkmax(T& a, const T& val) {return a < val ? a = val, 1 : 0;}
inline ull isqrt(ull k) {ull r = sqrt(k) + 1; while (r * r > k) r--; return r;}
inline ll icbrt(ll k) {ll r = cbrt(k) + 1; while (r * r * r > k) r--; return r;}
inline void addmod(int& a, int val, int p = MOD) {if ((a = (a + val)) >= p) a -= p;}
inline void submod(int& a, int val, int p = MOD) {if ((a = (a - val)) < 0) a += p;}
inline int mult(int a, int b, int p = MOD) {return (ll) a * b % p;}
inline int inv(int a, int p = MOD) {return fpow(a, p - 2, p);}
inline int sign(ld x) {return x < -EPS ? -1 : x > +EPS;}
inline int sign(ld x, ld y) {return sign(x - y);}
mt19937 mt(chrono::high_resolution_clock::now().time_since_epoch().count());
inline int mrand() {return abs((int) mt());}
inline int mrand(int k) {return abs((int) mt()) % k;}
#define db(x) cerr << "[" << #x << ": " << (x) << "] ";
#define endln cerr << "\n";

void chemthan() {
    string s; cin >> s;
    int n = sz(s);
    vector<vector<vi>> dp(n + 1, vector<vi>(n + 1, vi(n + 1)));
    dp[0][0][0] = 1;
    FOR(i, 0, n) {
        FOR(u, 0, n + 1) FOR(v, 0, n + 1) if (dp[i][u][v]) {
            if (i + 1 < n) {
                if (s[i] != s[i + 1]) {
                    dp[i + 2][u + 1][v] = 1;
                    dp[i + 2][u][v + 1] = 1;
                }
                else if (s[i] == '0') {
                    dp[i + 2][u + 1][v] = 1;
                }
                else {
                    dp[i + 2][u][v + 1] = 1;
                }
            }
            if (u) {
                dp[i + 1][u][v] = 1;
                if (s[i] != '0') {
                    dp[i + 1][u - 1][v + 1] = 1;
                }
            }
            if (v) {
                dp[i + 1][u][v] = 1;
                if (s[i] != '1') {
                    dp[i + 1][u + 1][v - 1] = 1;
                }
            }
        }
    }
    vector<vi> c(n + n + 1, vi(n + n + 1));
    FOR(i, 0, n + n + 1) c[0][i] = 1;
    FOR(i, 1, n + n + 1) FOR(j, 1, n + n + 1) c[i][j] = (c[i][j - 1] + c[i - 1][j - 1]) % MOD;
    vector<vi> f(n + 1, vi(n + 1));
    FOR(u, 0, n + 1) FOR(v, 0, n + 1) f[u][v] = c[u][u + v];
    static int g[333][333][333];
    int res = 0;
    FORd(i, n + 1, 2) {
        if (i < n) {
            vector<vi> nf(n + 1, vi(n + 1));
            auto f1 = f;
            auto f2 = f;
            FOR(i, 0, n + 1) FOR(j, 0, n + 1) {
                if (i) addmod(f1[i][j], f1[i - 1][j]);
                if (j) addmod(f2[i][j], f2[i][j - 1]);
            }
            FOR(u, 0, n + 1) FOR(v, 0, n + 1) {
                addmod(nf[u][v], f[u][v]);
                if (s[i] == '0') {
                    if (v) {
                        addmod(nf[u][v], f2[u][v - 1]);
                    }
                    //FOR(w, 0, v) {
                    //    addmod(nf[u][v], f[u][w]);
                    //}
                }
                else {
                    if (u) {
                        addmod(nf[u][v], f1[u - 1][v]);
                    }
                    //FOR(w, 0, u) {
                    //    addmod(nf[u][v], f[w][v]);
                    //}
                }
            }
            f = nf;
        }
        FOR(u, 0, n + 1) FOR(v, 0, n + 1) {
            int found = 0;
            if (i < n) {
                if (s[i] == '0' && g[i + 1][u + 1][v]) {
                    found = 1;
                }
                if (s[i] == '1' && g[i + 1][u][v + 1]) {
                    found = 1;
                }
            }
            if (found) {
                g[i][u][v] = 1;
            }
            if (!found && dp[i][u][v]) {
                addmod(res, f[u][v]);
                g[i][u][v] = 1;
            }
        }
    }
    addmod(res, 1);
    cout << res << "\n";
}

int main(int argc, char* argv[]) {
    ios_base::sync_with_stdio(0), cin.tie(0);
    if (argc > 1) {
        assert(freopen(argv[1], "r", stdin));
    }
    if (argc > 2) {
        assert(freopen(argv[2], "wb", stdout));
    }
    chemthan();
    cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
    return 0;
}

### Prompt

In cpp, your task is to solve the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
//#pragma GCC optimize("Ofast")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
 
#define ms(s, n) memset(s, n, sizeof(s))
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define FORd(i, a, b) for (int i = (a) - 1; i >= (b); --i)
#define FORall(it, a) for (__typeof((a).begin()) it = (a).begin(); it != (a).end(); it++)
#define sz(a) int((a).size())
#define present(t, x) (t.find(x) != t.end())
#define all(a) (a).begin(), (a).end()
#define uni(a) (a).erase(unique(all(a)), (a).end())
#define pb push_back
#define pf push_front
#define mp make_pair
#define fi first
#define se second
#define prec(n) fixed<<setprecision(n)
#define bit(n, i) (((n) >> (i)) & 1)
#define bitcount(n) __builtin_popcountll(n)
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pi;
typedef vector<int> vi;
typedef vector<pi> vii;
//const int MOD = (int) 1e9 + 7;
const int MOD = 119 << 23 | 1;
const int FFTMOD = 119 << 23 | 1;
const int INF = (int) 1e9 + 23111992;
const ll LINF = (ll) 1e18 + 23111992;
const ld PI = acos((ld) -1);
const ld EPS = 1e-9;
inline ll gcd(ll a, ll b) {ll r; while (b) {r = a % b; a = b; b = r;} return a;}
inline ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
inline ll fpow(ll n, ll k, int p = MOD) {ll r = 1; for (; k; k >>= 1) {if (k & 1) r = r * n % p; n = n * n % p;} return r;}
template<class T> inline int chkmin(T& a, const T& val) {return val < a ? a = val, 1 : 0;}
template<class T> inline int chkmax(T& a, const T& val) {return a < val ? a = val, 1 : 0;}
inline ull isqrt(ull k) {ull r = sqrt(k) + 1; while (r * r > k) r--; return r;}
inline ll icbrt(ll k) {ll r = cbrt(k) + 1; while (r * r * r > k) r--; return r;}
inline void addmod(int& a, int val, int p = MOD) {if ((a = (a + val)) >= p) a -= p;}
inline void submod(int& a, int val, int p = MOD) {if ((a = (a - val)) < 0) a += p;}
inline int mult(int a, int b, int p = MOD) {return (ll) a * b % p;}
inline int inv(int a, int p = MOD) {return fpow(a, p - 2, p);}
inline int sign(ld x) {return x < -EPS ? -1 : x > +EPS;}
inline int sign(ld x, ld y) {return sign(x - y);}
mt19937 mt(chrono::high_resolution_clock::now().time_since_epoch().count());
inline int mrand() {return abs((int) mt());}
inline int mrand(int k) {return abs((int) mt()) % k;}
#define db(x) cerr << "[" << #x << ": " << (x) << "] ";
#define endln cerr << "\n";

void chemthan() {
    string s; cin >> s;
    int n = sz(s);
    vector<vector<vi>> dp(n + 1, vector<vi>(n + 1, vi(n + 1)));
    dp[0][0][0] = 1;
    FOR(i, 0, n) {
        FOR(u, 0, n + 1) FOR(v, 0, n + 1) if (dp[i][u][v]) {
            if (i + 1 < n) {
                if (s[i] != s[i + 1]) {
                    dp[i + 2][u + 1][v] = 1;
                    dp[i + 2][u][v + 1] = 1;
                }
                else if (s[i] == '0') {
                    dp[i + 2][u + 1][v] = 1;
                }
                else {
                    dp[i + 2][u][v + 1] = 1;
                }
            }
            if (u) {
                dp[i + 1][u][v] = 1;
                if (s[i] != '0') {
                    dp[i + 1][u - 1][v + 1] = 1;
                }
            }
            if (v) {
                dp[i + 1][u][v] = 1;
                if (s[i] != '1') {
                    dp[i + 1][u + 1][v - 1] = 1;
                }
            }
        }
    }
    vector<vi> c(n + n + 1, vi(n + n + 1));
    FOR(i, 0, n + n + 1) c[0][i] = 1;
    FOR(i, 1, n + n + 1) FOR(j, 1, n + n + 1) c[i][j] = (c[i][j - 1] + c[i - 1][j - 1]) % MOD;
    vector<vi> f(n + 1, vi(n + 1));
    FOR(u, 0, n + 1) FOR(v, 0, n + 1) f[u][v] = c[u][u + v];
    static int g[333][333][333];
    int res = 0;
    FORd(i, n + 1, 2) {
        if (i < n) {
            vector<vi> nf(n + 1, vi(n + 1));
            auto f1 = f;
            auto f2 = f;
            FOR(i, 0, n + 1) FOR(j, 0, n + 1) {
                if (i) addmod(f1[i][j], f1[i - 1][j]);
                if (j) addmod(f2[i][j], f2[i][j - 1]);
            }
            FOR(u, 0, n + 1) FOR(v, 0, n + 1) {
                addmod(nf[u][v], f[u][v]);
                if (s[i] == '0') {
                    if (v) {
                        addmod(nf[u][v], f2[u][v - 1]);
                    }
                    //FOR(w, 0, v) {
                    //    addmod(nf[u][v], f[u][w]);
                    //}
                }
                else {
                    if (u) {
                        addmod(nf[u][v], f1[u - 1][v]);
                    }
                    //FOR(w, 0, u) {
                    //    addmod(nf[u][v], f[w][v]);
                    //}
                }
            }
            f = nf;
        }
        FOR(u, 0, n + 1) FOR(v, 0, n + 1) {
            int found = 0;
            if (i < n) {
                if (s[i] == '0' && g[i + 1][u + 1][v]) {
                    found = 1;
                }
                if (s[i] == '1' && g[i + 1][u][v + 1]) {
                    found = 1;
                }
            }
            if (found) {
                g[i][u][v] = 1;
            }
            if (!found && dp[i][u][v]) {
                addmod(res, f[u][v]);
                g[i][u][v] = 1;
            }
        }
    }
    addmod(res, 1);
    cout << res << "\n";
}

int main(int argc, char* argv[]) {
    ios_base::sync_with_stdio(0), cin.tie(0);
    if (argc > 1) {
        assert(freopen(argv[1], "r", stdin));
    }
    if (argc > 2) {
        assert(freopen(argv[2], "wb", stdout));
    }
    chemthan();
    cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
    return 0;
}

```