#include <bits/stdc++.h>
#if __has_include(<atcoder/all>)
#include <atcoder/all>
using namespace atcoder;
#endif
using namespace std;

/*
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
*/

#define ll long long
#define rep(i, n) for (ll i = 0; i < n; ++i)
#define rep_up(i, a, n) for (ll i = a; i < n; ++i)
#define rep_down(i, a, n) for (ll i = a; i >= n; --i)
#define P pair<ll, ll>
#define pb push_back
#define bit_count(x) __builtin_popcountll(x)
#define gcd(a,b) __gcd(a,b)
#define lcm(a,b) a / gcd(a,b) * b
#define endl "\n"

#define all(v) v.begin(), v.end()
#define fi first
#define se second
#define vvvvll vector< vector <vector <vector<ll> > > >
#define vvvll vector< vector< vector<ll> > >
#define vvll vector< vector<ll> >
#define vll vector<ll>
#define pqll priority_queue<ll>
#define pqllg priority_queue<ll, vector<ll>, greater<ll>>

template<class T> inline void vin(vector<T>& v) { rep(i, v.size()) cin >> v.at(i); }
template <class T>
using V = vector<T>;

constexpr ll INF = (1ll << 60);
//constexpr ll mod = 1000000007;
constexpr ll mod = 998244353;

constexpr double pi = 3.14159265358979323846;
template <typename T>
inline bool chmax(T &a, T b)
{
    if (a < b)
    {
        a = b;
        return 1;
    }
    return 0;
}
template <typename T>
inline bool chmin(T &a, T b)
{
    if (a > b)
    {
        a = b;
        return 1;
    }
    return 0;
}
template <typename T>
void pt(T val)
{
    cout << val << "\n";
}
template <typename T>
void pt_vll(vector<T> &v)
{
    ll vs = v.size();
    rep(i, vs)
    {
        cout << v[i];

        if (i == vs - 1)
            cout << "\n";
        else
            cout << " ";
    }
}

// O(N^2) 想定解

void solve(){
    ll n, m, k, cnt = 0, sum = 0, ans = 0;
    cin>>n;
    assert(1<=n&&n<=2000);
    string s;
    cin>>s;
    assert(s.size()==n);
    ll flag=1;
    rep(i,n){
        assert(s[i]=='0'||s[i]=='1'||s[i]=='?');
        if(s[i]=='0') flag=0;
    }
    ans+=flag;
    vvvll dp(n+1,vvll(2*n+3,vll(2)));
    // dp[i][j][k]:=i番目まで見て、(0の連続区間数*2 - 0の個数)+n+1=j, 最後の文字がk
    dp[0][n+1][1]=1;
    rep(i,n){
        rep(j,2*n+1){
            if(s[i]!='1'){
                dp[i+1][j][0]+=dp[i][j+1][0];
                dp[i+1][j+2][0]+=dp[i][j+1][1];
                dp[i+1][j][0]%=mod;
                dp[i+1][j+2][0]%=mod;
            }
            if(s[i]!='0'){
                dp[i+1][j+1][1]+=dp[i][j+1][0];
                dp[i+1][j+1][1]+=dp[i][j+1][1];
                dp[i+1][j+1][1]%=mod;
            }
        }
    }
    rep(i,2*n+3){
        if(i>=n+3&&(i+n+1)%2==0){
            ans+=dp[n][i][0];
            ans+=dp[n][i][1];
            ans%=mod;
        }
    }
    ans%=mod;
    pt(ans);
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    //cout << fixed << setprecision(16);
    //ll T;
    //cin>>T;
    //rep(ca,T)
    solve();
}