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:
#include <bits/stdc++.h>
// #include <iostream>
// #include <string>
// #include <vector>
// #include <queue>
// #include <algorithm>
// #include <assert.h>

using namespace std;
 
#define DEBUG(x) cerr<<#x<<": "<<x<<endl;
#define DEBUG_VEC(v) cerr<<#v<<":";for(int i=0;i<v.size();i++) cerr<<" "<<v[i]; cerr<<endl;
#define DEBUG_MAT(v) cerr<<#v<<endl;for(int i=0;i<v.size();i++){for(int j=0;j<v[i].size();j++) {cerr<<v[i][j]<<" ";}cerr<<endl;}
typedef long long ll;
// #define int ll
 
#define vi vector<int>
#define vl vector<ll>
#define vii vector< vector<int> >
#define vll vector< vector<ll> >
#define vs vector<string>
#define pii pair<int,int>
#define pis pair<int,string>
#define psi pair<string,int>
#define pll pair<ll,ll>
template<class S, class T> pair<S, T> operator+(const pair<S, T> &s, const pair<S, T> &t) { return pair<S, T>(s.first + t.first, s.second + t.second); }
template<class S, class T> pair<S, T> operator-(const pair<S, T> &s, const pair<S, T> &t) { return pair<S, T>(s.first - t.first, s.second - t.second); }
template<class S, class T> ostream& operator<<(ostream& os, pair<S, T> p) { os << "(" << p.first << ", " << p.second << ")"; return os; }
#define X first
#define Y second
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)
#define rrep(i,n) for(int i=(int)(n)-1;i>=0;i--)
#define rrep1(i,n) for(int i=(int)(n);i>0;i--)
#define REP(i,a,b) for(int i=a;i<b;i++)
#define in(x, a, b) (a <= x && x < b)
#define all(c) c.begin(),c.end()
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a = b; return 1; } return 0; }
#define UNIQUE(v) v.erase(std::unique(v.begin(), v.end()), v.end());
const ll inf = 1000000001;
const ll INF = (ll)1e18 + 1;
const long double pi = 3.1415926535897932384626433832795028841971L;
#define Sp(p) cout<<setprecision(25)<< fixed<<p<<endl;
// int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
// int dx2[8] = { 1,1,0,-1,-1,-1,0,1 }, dy2[8] = { 0,1,1,1,0,-1,-1,-1 };
// vi dx = {1, 0, -1, 0}, dy = {0, 1, 0, -1};
// vi dx2 = { 1,1,0,-1,-1,-1,0,1 }, dy2 = { 0,1,1,1,0,-1,-1,-1 };
#define fio() cin.tie(0); ios::sync_with_stdio(false);
// const ll MOD = 1000000007;
const ll MOD = 998244353;
// #define mp make_pair
//#define endl '\n'



signed main() {
    fio();
    string s;
    cin >> s;
    int n = s.size();
    vector<vector<vector<bool>>> dp(n + 1, vector<vector<bool>>(n + 1, vector<bool>(n + 1)));
    dp[0][0][0] = true;
    rep (i, n) {
        rrep (j, n + 1) {
            rrep (k, n + 1) {
                if (not dp[i][j][k]) continue;
                if (i + 2 <= n) {
                    dp[i + 1][j][k] = true;
                    if (s[i] == '0') dp[i + 2][j + 1][k] = true;
                    else dp[i + 2][j][k + 1] = true;
                    if (s[i + 1] == '0') dp[i + 2][j + 1][k] = true;
                    else dp[i + 2][j][k + 1] = true;
                }

                if (j + k >= 2 and j > 0) dp[i][j - 1][k] = true;
                if (j + k >= 2 and k > 0) dp[i][j][k - 1] = true;

                if (j + k >= 1) dp[i + 1][j][k] = true;
                if (s[i] == '0' and k > 0) dp[i + 1][j + 1][k - 1] = true;
                if (s[i] == '1' and j > 0) dp[i + 1][j - 1][k + 1] = true;
            }
        }
    }

    reverse(all(s));
    vector<vll> dp2(n + 1, vll(n + 1, vl(n + 1)));
    dp2[0][0][0] = 1;
    rep (i, n) {
        rep (j, n + 1) {
            rep (k, n + 1) {
                if (dp2[i][j][k] == 0) continue;
                // 次に作りたいのが0の時
                if (s[i] == '0') (dp2[i + 1][j][k] += dp2[i][j][k]) %= MOD;
                else {
                    if (j + 1 <= n) (dp2[i][j + 1][k] += dp2[i][j][k]) %= MOD;
                }
                // 次に作りたいのが1の時
                if (s[i] == '1') (dp2[i + 1][j][k] += dp2[i][j][k]) %= MOD;
                else {
                    if (k + 1 <= n) (dp2[i][j][k + 1] += dp2[i][j][k]) %= MOD;
                }
            }
        }
    }

    ll ans = 0;
    for (int i = 0; i <= n; i++) {
        for (int j = 0; j <= n; j++) {
            for (int k = 0; k <= n; k++) {
                if (i == n and j == 0 and k == 0) continue;
                if (dp[i][j][k]) {
                    (ans += dp2[n - i][j][k]) %= MOD;
                }
            }
        }
    }
    cout << ans << endl;
}