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<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<cassert>
#include<complex>
#include<numeric>
using namespace std;

//#define int long long
typedef long long ll;

typedef unsigned long long ul;
typedef unsigned int ui;
constexpr ll mod = 998244353;
const ll INF = mod * mod;
typedef pair<int, int>P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
#define all(v) (v).begin(),(v).end()
typedef pair<ll, ll> LP;
typedef long double ld;
typedef pair<ld, ld> LDP;
const ld eps = 1e-12;
const ld pi = acos(-1.0);

ll mod_pow(ll x, ll n) {
	ll res = 1;
	while (n) {
		if (n & 1)res = res * x%mod;
		x = x * x%mod; n >>= 1;
	}
	return res;
}
struct modint {
	ll n;
	modint() :n(0) { ; }
	modint(ll m) :n(m) {
		if (n >= mod)n %= mod;
		else if (n < 0)n = (n%mod + mod) % mod;
	}
	operator int() { return n; }
};
bool operator==(modint a, modint b) { return a.n == b.n; }
modint operator+=(modint &a, modint b) { a.n += b.n; if (a.n >= mod)a.n -= mod; return a; }
modint operator-=(modint &a, modint b) { a.n -= b.n; if (a.n < 0)a.n += mod; return a; }
modint operator*=(modint &a, modint b) { a.n = ((ll)a.n*b.n) % mod; return a; }
modint operator+(modint a, modint b) { return a += b; }
modint operator-(modint a, modint b) { return a -= b; }
modint operator*(modint a, modint b) { return a *= b; }
modint operator^(modint a, int n) {
	if (n == 0)return modint(1);
	modint res = (a*a) ^ (n / 2);
	if (n % 2)res = res * a;
	return res;
}

ll inv(ll a, ll p) {
	return (a == 1 ? 1 : (1 - p * inv(p%a, a)) / a + p);
}
modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }

const int max_n = 1 << 20;
modint fact[max_n], factinv[max_n];
void init_f() {
	fact[0] = modint(1);
	for (int i = 0; i < max_n - 1; i++) {
		fact[i + 1] = fact[i] * modint(i + 1);
	}
	factinv[max_n - 1] = modint(1) / fact[max_n - 1];
	for (int i = max_n - 2; i >= 0; i--) {
		factinv[i] = factinv[i + 1] * modint(i + 1);
	}
}
modint comb(int a, int b) {
	if (a < 0 || b < 0 || a < b)return 0;
	return fact[a] * factinv[b] * factinv[a - b];
}


modint dp[303][303][303];
modint dp2[303][303][303];

//i回捨て、そのうちj回が0である時の右端
int madp[303][303];
void solve() {
	string s; cin >> s;
	int n = s.size();
	dp[0][0][0] = 1;
	dp2[0][0][0] = 1;
	rep(i, n + 1) {
		rep(j, n + 1) {
			rep(k, n + 1) {
				//put 0
				if (j < n&&s[n - 1 - j] == '0') {
					dp[i + 1][j + 1][k] += dp[i][j][k];
					dp2[i + 1][j + 1][k] += dp[i][j][k];
				}
				else {
					dp[i + 1][j][k + 1] += dp[i][j][k];
				}
				//put 1
				if (j < n&&s[n - 1 - j] == '1') {
					dp[i + 1][j + 1][k] += dp[i][j][k];
					dp2[i + 1][j + 1][k] += dp[i][j][k];
				}
				else {
					dp[i + 1][j][k] += dp[i][j][k];
				}

				//put 0
				dp2[i + 1][j][k + 1] += dp2[i][j][k];
				//put 1
				dp2[i + 1][j][k] += dp2[i][j][k];
			}
		}
	}

	vector<int> cnt(n + 1);
	rep(i, n) {
		cnt[i + 1] += cnt[i];
		if (s[i] == '0')cnt[i + 1]++;
	}
	rep(i, n + 1)rep(j, n + 1)madp[i][j] = -mod;
	madp[0][0] = 0;
	rep(i, n-1) {
		rep(j, i+1) {
			if (madp[i][j] < 0)continue;
			
			//take new x
			rep(x, 3) {
				int ri = madp[i][j];
				if (ri + x <= n) {
					int c0 = cnt[ri + x] - j;
					int c1 = ri + x - i - c0;
					if (c0 < 0 || c1 < 0)continue;
					if (c0 > 0) {
						if (x == 2 && s[ri] == '1'&&s[ri + 1] == '1') {
							//
						}
						else 
							madp[i + 1][j + 1] = max(madp[i + 1][j + 1], ri + x);
					}
					if (c1 > 0) {
						if (x == 2 && s[ri] == '0'&&s[ri + 1] == '0') {
							//
						}
						else
						madp[i + 1][j] = max(madp[i + 1][j], ri + x);
					}
				}
			}
		}
	}

	modint ans = 0;
	rep(i, n) {
		rep(j, i + 1) {
			if (madp[i][j] < 0)continue;
			int num = n - madp[i][j];
			int nnum = cnt[madp[i][j]] - j;
			ans += dp2[n - i][num][nnum];
			//cout << n-i << " " << num << " " << nnum <<" "<<dp2[n-i][num][nnum] <<"\n";
		}
	}

	cout << ans << "\n";
}


signed main() {
	//ios::sync_with_stdio(false);
	//cin.tie(0);
	//cout << fixed << setprecision(10);
	init_f();
	//int t; cin >> t; rep(i, t)
		solve();
	stop
		return 0;
}