#include<bits/stdc++.h>
//#pragma GCC optimize(3)
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define int long long
#define ll long long
#define ull unsigned ll
#define db double
#define ldb long db
#define mp make_pair
#define fi first
#define se second
#define pii pair<int,int>
#define pbI push_back
#define inf 1e9
#define mdI int mid=(l+r)>>1
#define lowbit(x) (x&-x)
#define rep(a,b,c) for(int a=b;a<=c;a++)
#define MAXN 520
#define gint __int128
#define MOD 998244353
int qread(){
	char o=getchar();int f=1,x=0;
	for(;!isdigit(o);o=getchar())if(o=='-')f*=-1;
	for(;isdigit(o);o=getchar())x=x*10+o-'0';
	return f*x;
}
void chmx(int &x,int y){if(y>x)x=y;}
void chmi(int &x,int y){if(y<x)x=y;}
bool FLA;
namespace MathPart{
	#define rep(a,b,c) for(int a=b;a<=c;a++)
	#define vep(a,b,c) for(int a=b;a>=c;a--)
	void exgcd(ll a,ll b,ll &x,ll &y){if(b==0){x=1,y=0;return;}exgcd(b,a%b,y,x);y-=(a/b)*x;}
	ll inv(ll a,ll p=MOD){ll x,y;exgcd(a,p,x,y);return (x+p)%p;}
	gint qp(ll a,ll b,ll p=MOD){gint rs=1,bs=a;while(b){if(b&1){rs=rs*bs%p;}bs=bs*bs%p;b>>=1;}return rs;}
	ll fac[MAXN],ifac[MAXN];
	void initF(ll n,ll p=MOD){
		fac[0]=1;rep(i,1,n){fac[i]=fac[i-1]*i%p;}
		ifac[n]=inv(fac[n],p);vep(i,n-1,0){ifac[i]=ifac[i+1]*(i+1)%p;}
	}
	ll C(ll n,ll m){if(n<m){return 0;}return fac[n]*ifac[m]%MOD*ifac[n-m]%MOD;}
	ll CC(ll n,ll m,ll p=MOD){ll rs=1;rep(i,0,m-1){rs=rs*(n-i)%p*inv(i+1,MOD)%p;}return rs;}
	ll S2OL(ll n,ll m){
		ll rs=0,d=1,f=1;
		for(ll k=0;k<=m;k++){
			rs=(rs+f*C(m,k)*qp(m-k,n,MOD)%MOD+MOD)%MOD;
			d=d*max(k,1ll)%MOD;f*=-1;
		}return rs*inv(d,MOD)%MOD;
	}
	ll phi(ll a){ll rs=a,m=sqrt(a);
		rep(i,2,m){if(a%i==0){rs=rs/i*(i-1);while(a%i==0){a/=i;}}}
		if(a>1){rs=rs/a*(a-1);}return rs;
	}
	bool MR(ll pr){if(pr<2)return 0;if(pr==3)return 1;if(pr==2)return 1;
		ll a=pr-1,b=0;while(!(a&1))a>>=1,b++;
		for(int i=1;i<=8;++i){
		    ll x=rand()%(pr-2)+2,v=qp(x,a,pr),j;if(v==1||v==pr-1)continue;
		    for(j=0;j<b-1;j++){v=(gint)v*v%pr;if(v==pr-1)break;}
			if(v!=pr-1)return 0;
		}return 1;
	}
}
//using namespace MathPart;
int dp[2][MAXN*2][MAXN*2][2];//i s_i max_{j=1}^i(s_i) [s_i=0/1]
int off=500;
int n;string S;
bool FLB;
signed main(){
	cerr<<((&FLB)-(&FLA))/1024.0/1024<<endl;
	cin>>n;cin>>S;
	reverse(S.begin(),S.end());S=" "+S;
	dp[0][off][off][0]=1;
	rep(i,0,n-1){
        bool s=i&1;
        memset(dp[!s],0,sizeof(dp[!s]));
        rep(j,0,off+off){
            rep(k,j,off+off){
                if(S[i+1]=='0'||S[i+1]=='?'){
                    if(j){
                        dp[!s][j-1][k][0]+=dp[s][j][k][0]+dp[s][j][k][1];
                        dp[!s][j][k][0]%=MOD;
                    }
                }
                if(S[i+1]=='1'||S[i+1]=='?'){
                    if(j+3<=2*off&&k>=j+2){
                        dp[!s][j+3][max(k,j+3)][1]+=dp[s][j][k][0];
                        dp[!s][j+3][max(k,j+3)][1]%=MOD;
                    }
                }
            }
        }
	}
	int ans=0;
	rep(i,0,2*off){ans+=dp[n&1][off][i][0];ans%=MOD;}cout<<ans;
}
/*
8
0??0?100
由判定性问题展开
处理01序列的经典方法:设定数码的权值，直观折线图
(找到对目前判定方法的反例:0010问题)
下文s表示前缀和
0=-1,1=+3
对于一个1他不能匹配因为他右边下不去,
翻转一下,左边下不去导致的，
充要条件:无相邻1,最后非1,和=0,每一个i处的1，存在一个j使得s_j=s_i-1(知其下降过程连续,实际上和最大值等价)
必要性:递归构造可得
充分性:取第一个最大的s_i(一定为1),由条件知s_{i-1}=s_i-3,s_{i-2}=s_i-2,找到如是j(j最小),s_{i+1}=s_i-1,s_{i+2}=s_i-2(最大性)
观察[j+1,i-1]平的,j+2向下(否则没有最小性),找到(j+1,j+2,i,i+1)这一组，随后可以删除、拼接、归纳


跳出区间dp套路
*/