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:
/* Aa^~ kokoro ga pyonpyon suru n jaa^~
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// ggmgmmqqqH$ ``````````....````` `  ````````.`````` `` ``````.yfpppbbbbkkkkqqqqqH
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// @@g@gggHH=.`..._<-___..```````````````````````. .-_~~~_(!``-.``.`` OHHWUWHmqHWXW
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// kkkk._(_.->..._(z;:_><.._>_+_<(1>_._<...(v<<.(<.(+z<..-_(Z~_<_j+_..`...`(WHKz1ZW
// @@gR._+_..~..-<+z<<?<>```_.<_.(+1><_;_..(1_:`.<<??1z--(+Z!..<_.j<....`..(bgHAAQX
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// @gmH_(zl..(.._+>```<+_````.~>``(+.<?>>_._(<```(<<``(__<>....<.._<.......dXkkkHHH
// mmqHl(dk_.(_.-=~`.`.1-..._~-1.``_:`(??<_~(`.--.&_.`.<(;<...._.._<..`..._Xg@@@@@@
// qHkpk(dX<.(;..j_```.(((JJ&a&-~``````.1<_```-(((e+.-(/`(>...._..(<......(Wmggg@@g
// HVHbWcz><__+_.(_.(dWWHHH@HHc~````````.+~`` (jHMMMHHHm&.?..._<..(<_..._.(WqqHHmHg
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// X0XWHKXXw$<(z<.( `` WHHMHHHH_``````````````.WHHMNMHHH_`(...(<_.(z_..._<(fWVC174W
// XuXWHHWWz>__+z+.!`..??CZYCOX_`````````````.`~.OvTUZUS_`~.._+?_.(_~_.._zjO=1+~+jy
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// HHHmHSZu:(_~+OztttXtttOZZttO+-..............-(+ztOttwttttd0tOZttttwOl<~.(_dMMHHH
// rvuuXuuI~~<~(uttttwvOwwwkQQHMMHHHHHHHHHMMMNmgey?OwwwrtttwXOtwttttttXtO-~.((wZyyy
// HHHHHHK>(~(-(dOrtrrl(QgMHMMMHHHHHHHHHHHHHHHH##HMNkX0rrrrXXrd%`` (Ctwwtz_~.<(Wg@H
// NNNNNHD(~(zo~zXrrrQdHHMMNMHHHHHHHHHHHHHHHHHHHHHH##HNmyrdKkwZ ` _``-zwrt1~~_<(MNM
// MMMMM#<<_jwr:(Z4QHHMMHMHHHHHHHHHHHHHHHHHHHHHHHHHHHH###NHSXZ>` ~````.OXtt>~._<?MM
*/
#include<bits/stdc++.h>
using namespace std;
#ifndef ONLINE_JUDGE
#define dbg(x...) do { cout << "\033[32;1m " << #x << " -> "; err(x); } while (0)
void err() { cout << "\033[39;0m" << endl; }
template<template<typename...> class T, typename t, typename... A>
void err(T<t> a, A... x) { for (auto v: a) cout << v << ' '; err(x...); }
template<typename T, typename... A>
void err(T a, A... x) { cout << a << ' '; err(x...); }
#else
#define dbg(...)
#endif
typedef long long ll;
typedef pair<int,int> pi;
typedef vector<int> vi;
template<class T> using vc=vector<T>;
template<class T> using vvc=vc<vc<T>>;
template<class T> void mkuni(vector<T>&v)
{
    sort(v.begin(),v.end());
    v.erase(unique(v.begin(),v.end()),v.end());
}
ll rand_int(ll l, ll r) //[l, r]
{
    static mt19937_64 gen(chrono::steady_clock::now().time_since_epoch().count());
    return uniform_int_distribution<ll>(l, r)(gen);
}
template<class T>
void print(T x,int suc=1)
{
    cout<<x;
    if(suc==1) cout<<'\n';
    else cout<<' ';
}
template<class T>
void print(const vector<T>&v,int suc=1)
{
    for(int i=0;i<v.size();i++)
    print(v[i],i==(int)(v.size())-1?suc:2);
}
const int maxn=305;
const int mod=998244353;
int dp[maxn][maxn][maxn];
bool ok[maxn][maxn][maxn];
int C[maxn][maxn];
void init()
{
    for(int i=0;i<maxn;i++)
    {
        C[i][0]=1;
        for(int j=1;j<=i;j++)
            C[i][j]=(C[i-1][j-1]+C[i-1][j])%mod;
    }
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    init();
    string s;
    cin>>s;
    int n=s.length();
    for(int j=0;j<=n;j++)
    {
        for(int k=0;k+j<=n;k++)
            dp[n][j][k]=C[k+j][k];
    }
    for(int i=n-1;i>=0;i--)
    {
        for(int j=0;j<=i;j++)
        {
            for(int k=0;k+j<=i;k++)
            {
                dp[i][j][k]=(dp[i][j][k]+dp[i+1][j][k])%mod;
                if(s[i]=='0'&&k)
                    dp[i][j][k]=(dp[i][j][k-1]+dp[i][j][k])%mod;
                if(s[i]=='1'&&j)
                    dp[i][j][k]=(dp[i][j-1][k]+dp[i][j][k])%mod;
                // dbg(i,j,k,dp[i][j][k]);
            }
        }
    }
    // for()
    int ans=0;
    ok[0][0][0]=1;
    for(int i=1;i<=n;i++)
    {
        for(int j=i;j>=0;j--)
        {
            for(int k=i-j;k>=0;k--)
            {
                if(j+k)
                {
                    ok[i][j][k]|=(ok[i][j+1][k]|ok[i][j][k+1]);// 自交
                    ok[i][j][k]|=ok[i-1][j][k];// 删除当前字符
                }
                if(s[i-1]=='0'&&j)
                    ok[i][j][k]|=ok[i-1][j-1][k+1];
                if(s[i-1]=='1'&&k)
                    ok[i][j][k]|=ok[i-1][j+1][k-1];
                // 两个字符移进
                if(i>=2)
                {
                    if(j)
                    {
                        if(s[i-1]=='0'||s[i-2]=='0') ok[i][j][k]|=ok[i-2][j-1][k];
                    }
                    if(k)
                    {
                        if(s[i-1]=='1'||s[i-2]=='1') ok[i][j][k]|=ok[i-2][j][k-1];
                    }
                }
                // dbg(i,j,k,ok[i][j][k]);
            }
        }
    }
    for(int i=n;i>=0;i--)
    {
        for(int j=0;j<=i;j++)
        {
            for(int k=0;k+j<=i;k++)
            {
                if(ok[i][j][k])
                {
                    ans=(ans+dp[i][j][k])%mod;
                    vi cnt(2);
                    for(int l=i-1;l>=0;l--)// 保证不重复
                    {
                        cnt[s[l]=='1']++;
                        if(cnt[0]>j||cnt[1]>k) break;
                        ok[l][j-cnt[0]][k-cnt[1]]=false;
                    }
                    // dbg(i,j,k,dp[i][j][k]);
                }
            }
        }
    }
    print(ans);

}