#include<bits/stdc++.h>
using namespace std;

//* ATCODER
#include<atcoder/all>
using namespace atcoder;
typedef modint998244353 mint;
//*/

/* BOOST MULTIPRECISION
#include<boost/multiprecision/cpp_int.hpp>
using namespace boost::multiprecision;
//*/

typedef long long ll;

#define rep(i, s, n) for (int i = (int)(s); i < (int)(n); i++)
#define rrep(i, s, n) for (int i = (int)(n)-1; i >= (int)(s); i--)

template <typename T> bool chmin(T &a, const T &b) {
	if (a <= b) return false;
	a = b;
	return true;
}

template <typename T> bool chmax(T &a, const T &b) {
	if (a >= b) return false;
	a = b;
	return true;
}

template <typename T> T max(vector<T> &a){
	assert(!a.empty());
	T ret = a[0];
	for (int i=0; i<(int)a.size(); i++) chmax(ret, a[i]);
	return ret;
}

template <typename T> T min(vector<T> &a){
	assert(!a.empty());
	T ret = a[0];
	for (int i=0; i<(int)a.size(); i++) chmin(ret, a[i]);
	return ret;
}

template <typename T> T sum(vector<T> &a){
	T ret = 0;
	for (int i=0; i<(int)a.size(); i++) ret += a[i];
	return ret;
}

//defmodfact
const int COMinitMAX = 1998244;
mint fact[COMinitMAX+1], factinv[COMinitMAX+1];

void modfact(){
	fact[0] = 1;
	for (int i=1; i<=COMinitMAX; i++){
		fact[i] = fact[i-1] * i;
	}
	factinv[COMinitMAX] = fact[COMinitMAX].inv();
	for (int i=COMinitMAX-1; i>=0; i--){
		factinv[i] = factinv[i+1] * (i+1);
	}
}

mint binom(int a, int b){
	if (a<b || b<0) return mint(0);
	return fact[a]*factinv[b]*factinv[a-b];
}
//--------


int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	modfact();
	
	int n; cin >> n;
	string s; cin >> s;
	vector<int> a(n);
	rep(i,0,n){
		a[i] = (s[i] == '1');
	}

	vector dp(n+1, vector<mint>(n+1));
	dp[0][0] = 1;
	// 下に何個あるか ?

	rep(i,0,n){
		if (a[i] == 0){
			// 下から取る.
			rep(j,1,n+1){
				if (n-(j+i) < 0) break;
				dp[i+1][j-1] += dp[i][j] * j;
			}
			// 上を選ぶ
			rep(j,0,n+1){
				if (n-(j+i) < 0) break;
				rep(k,1,n-(j+i)+1){
					dp[i+1][k+j-1] += dp[i][j];
				}
			}
		}else{
			// 上を選ぶ
			rep(j,0,n+1){
				if (n-(j+i) < 0) break;
				dp[i+1][j] += dp[i][j] * (n-(j+i));
			}
		}
	}
	cout << sum(dp[n]).val() << '\n';
		
}