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:
// #pragma GCC target("avx")  // CPU 処理並列化
// #pragma GCC optimize("O3")  // CPU 処理並列化
// #pragma GCC optimize("unroll-loops")  // 条件処理の呼び出しを減らす
// #define BEGIN_STACK_EXTEND(size) void * stack_extend_memory_ = malloc(size);void * stack_extend_origin_memory_;char * stack_extend_dummy_memory_ = (char*)alloca((1+(int)(((long long)stack_extend_memory_)&127))*16);*stack_extend_dummy_memory_ = 0;asm volatile("mov %%rsp, %%rbx\nmov %%rax, %%rsp":"=b"(stack_extend_origin_memory_):"a"((char*)stack_extend_memory_+(size)-1024));
// #define END_STACK_EXTEND asm volatile("mov %%rax, %%rsp"::"a"(stack_extend_origin_memory_));free(stack_extend_memory_);

#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<deque>
#include<stack>
#include<string>
#include<string.h>
#include<vector>
#include<set>
#include<map>
#include<bitset>
#include<stdlib.h>
#include<cassert>
#include<time.h>
#include<bitset>
#include<numeric>
#include<unordered_set>
#include<unordered_map>
#include<complex>
using namespace std;
const long long mod=998244353;
const long long inf=mod*mod;
const long long d2=(mod+1)/2;
const double EPS=1e-11;
const double INF=1e+10;
const double PI=acos(-1.0);
const int C_SIZE = 11100000;
const int UF_SIZE = 3100000;
namespace{
	long long fact[C_SIZE];
	long long finv[C_SIZE];
	long long inv[C_SIZE];
	inline long long Comb(int a,int b){
	 	if(a<b||b<0)return 0;
	 	return fact[a]*finv[b]%mod*finv[a-b]%mod;
	}
	void init_C(int n){
		fact[0]=finv[0]=inv[1]=1;
		for(int i=2;i<n;i++){
			inv[i]=(mod-(mod/i)*inv[mod%i]%mod)%mod;
		}
		for(int i=1;i<n;i++){
			fact[i]=fact[i-1]*i%mod;
			finv[i]=finv[i-1]*inv[i]%mod;
		}
	}
	long long pw(long long a,long long b){
		if(a<0LL)return 0;
		if(b<0LL)return 0;
		long long ret=1;
		while(b){
			if(b%2)ret=ret*a%mod;
			a=a*a%mod;
			b/=2;
		}
		return ret;
	}
	long long pw_mod(long long a,long long b,long long M){
		if(a<0LL)return 0;
		if(b<0LL)return 0;
		long long ret=1;
		while(b){
			if(b%2)ret=ret*a%M;
			a=a*a%M;
			b/=2;
		}
		return ret;
	}
	int pw_mod_int(int a,int b,int M){
		if(a<0)return 0;
		if(b<0)return 0;
		int ret=1;
		while(b){
			if(b%2)ret=(long long)ret*a%M;
			a=(long long)a*a%M;
			b/=2;
		}
		return ret;
	}
	int ABS(int a){return max(a,-a);}
	long long ABS(long long a){return max(a,-a);}
	double ABS(double a){return max(a,-a);}
	int sig(double r) { return (r < -EPS) ? -1 : (r > +EPS) ? +1 : 0; }
	int UF[UF_SIZE];
	void init_UF(int n){
		for(int i=0;i<n;i++)UF[i]=-1;
	}
	int FIND(int a){
		if(UF[a]<0)return a;
		return UF[a]=FIND(UF[a]);
	}
	void UNION(int a,int b){
		a=FIND(a);b=FIND(b);if(a==b)return;
		if(UF[a]>UF[b])swap(a,b);
		UF[a]+=UF[b];UF[b]=a;
	}
}
// ここから編集しろ
char in[310];
long long dp[310][310][310];
int req[310][310];
int ijk[310][310][310];
int v[310][310][310];
char C[310];
int main(){
	scanf("%s",in);
	int n=strlen(in);
	for(int i=0;i<n;i++)C[i]=in[i];
	reverse(C,C+n);
	
	for(int i=0;i<310;i++)for(int j=0;j<310;j++)for(int k=0;k<310;k++)
		ijk[i][j][k]=mod;
	ijk[0][0][0]=0;
	priority_queue<pair<pair<int,int>,pair<int,int> > > Q;
	Q.push(make_pair(make_pair(0,0),make_pair(0,0)));
	while(Q.size()){
		int cost=-Q.top().first.first;
		int C=Q.top().first.second;
		int A=Q.top().second.first;
		int B=Q.top().second.second;
		Q.pop();
		if(v[A][B][C])continue;
		v[A][B][C]=1;
		int at=cost;
		
		if(at+2<=n){
			if(ijk[A][B][C+1]>cost+2){
				ijk[A][B][C+1]=cost+2;
				Q.push(make_pair(make_pair(-cost-2,C+1),make_pair(A,B)));
			}
			if(in[at]=='0'||in[at+1]=='0'){
				if(ijk[A+1][B][C]>cost+2){
					ijk[A+1][B][C]=cost+2;
					Q.push(make_pair(make_pair(-cost-2,C),make_pair(A+1,B)));
				}
			}
			if(in[at]=='1'||in[at+1]=='1'){
				if(ijk[A][B+1][C]>cost+2){
					ijk[A][B+1][C]=cost+2;
					Q.push(make_pair(make_pair(-cost-2,C),make_pair(A,B+1)));
				}
			}
		}
		if(at+1<=n&&C){
			if(ijk[A][B][C-1]>cost+1){
				ijk[A][B][C-1]=cost+1;
				Q.push(make_pair(make_pair(-cost-1,C-1),make_pair(A,B)));
			}
			if(in[at]=='0'){
				if(ijk[A+1][B][C-1]>cost+1){
					ijk[A+1][B][C-1]=cost+1;
					Q.push(make_pair(make_pair(-cost-1,C-1),make_pair(A+1,B)));
				}
			}
			if(in[at]=='1'){
				if(ijk[A][B+1][C-1]>cost+1){
					ijk[A][B+1][C-1]=cost+1;
					Q.push(make_pair(make_pair(-cost-1,C-1),make_pair(A,B+1)));
				}
			}
		}
	}
	for(int i=0;i<=n;i++){
		for(int j=0;j<=n;j++){
			req[i][j]=mod;
			for(int k=0;k<=n;k++){
				req[i][j]=min(req[i][j],ijk[i][j][k]);
			}
		}
	}
	dp[0][0][0]=1;
	for(int i=0;i<n;i++){
		for(int j=0;j<=n;j++){
			for(int k=0;k<=n;k++){
				if(!dp[i][j][k])continue;
				if(req[j][k]+i<n){
					dp[i+1][j][k]=(dp[i+1][j][k]+dp[i][j][k])%mod;
				}
				if(C[i]=='1'){
					if(req[j+1][k]+i<=n){
						dp[i][j+1][k]=(dp[i][j+1][k]+dp[i][j][k])%mod;
					}
				}else{
					if(req[j][k+1]+i<=n){
						dp[i][j][k+1]=(dp[i][j][k+1]+dp[i][j][k])%mod;
					}
				}
			}
		}
	}
	long long ret=0;
	for(int i=0;i<=n;i++)for(int j=0;j<=n;j++)for(int k=0;k<=n;k++){
		ret=(ret+dp[i][j][k])%mod;
	}
	ret=(ret+mod-1)%mod;
	printf("%lld\n",ret);
}