#include <iostream>
#include <vector>
#include <cmath>
#include <map>
#include <set>
#include <iomanip>
#include <queue>
#include <algorithm>
#include <numeric>
#include <deque>
#include <complex>
#include <cassert>

using namespace std;
using ll = long long;

const ll modc = 9982344353;
class mint {
    ll x;
public:
    mint(ll x=0) : x((x%modc+modc)%modc) {}
    mint operator-() const {
      return mint(-x);
    }
    mint& operator+=(const mint& a) {
        if ((x += a.x) >= modc) x -= modc;
        return *this;
    }
    mint& operator-=(const mint& a) {
        if ((x += modc-a.x) >= modc) x -= modc;
        return *this;
    }
    mint& operator*=(const  mint& a) {
        (x *= a.x) %= modc;
        return *this;
    }
    mint operator+(const mint& a) const {
        mint res(*this);
        return res+=a;
    }
    mint operator-(const mint& a) const {
        mint res(*this);
        return res-=a;
    }
    mint operator*(const mint& a) const {
        mint res(*this);
        return res*=a;
    }
    mint pow(ll t) const {
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }
    mint inv() const {
        return pow(modc-2);
    }
    mint& operator/=(const mint& a) {
        return (*this) *= a.inv();
    }
    mint operator/(const mint& a) const {
        mint res(*this);
        return res/=a;
    }
    bool operator == (const mint& a) const{
        return x == a.x;
    }
    friend ostream& operator<<(ostream& os, const mint& m){
        os << m.x;
        return os;
    }
    friend istream& operator>>(istream& ip, mint&m) {
        ip >> m.x;
        return ip;
    }
};

ll lucas(ll N, ll K){
    return ((N & K) == K ? 1 : 0);
}

int main(){

    int N;
    cin >> N;
    vector<int> A(N);
    for (int i=0; i<N; i++) cin >> A[i];
    vector<vector<mint>> dp(N+1, vector<mint>(2));
    dp[0][0] = 1;
    for (int i=1; i<=N; i++){
        for (int j=0; j<2; j++){
            if (A[i-1] != -1){
                if (lucas(N-1, i-1)) dp[i][j^A[i-1]] += dp[i-1][j];
                else dp[i][j] += dp[i-1][j];
            }
            else{
                if (lucas(N-1, i-1)) dp[i][j] += dp[i-1][0] + dp[i-1][1];
                else dp[i][j] += dp[i-1][j] * 2;
            }
        }
    }

    cout << dp[N][1] << endl;

    return 0;
}