#include <bits/stdc++.h>
#include <atcoder/all>
using namespace std;
using namespace atcoder;
using ll = long long;
using ld = long double;
using ull = unsigned long long;
using Graph = vector<vector<int>>;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define all(v) (v).begin(), (v).end()
#define rall(v) (v).rbegin(), (v).rend()
#define NP next_permutation
const int INF32 = 2e9;
const ll INF64 = 2e18;
const ll mod = 998244353;
// const ll mod = 1e9 + 7;
template<typename T> bool chmin(T &a, T b){if(a > b){a = b; return true;} return false;}
template<typename T> bool chmax(T &a, T b){if(a < b){a = b; return true;} return false;}
 
void cincout() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout << fixed << setprecision(15);
}
 
int main() {
    cincout();
 
    int N;
    cin >> N;

    vector<int> c(N);
    for (int i = 0; i < N; i++) cin >> c[i];

    if (N == 1) {
        cout << 1 << endl;
        return 0;
    } 

    using mint = modint998244353;
    mint ans = 1;
    if (c[0] < 0 && 0 < c[N - 1]) {
        int id = lower_bound(all(c), 0) - c.begin();
        if (c[id] == 0) {
            ans *= 2 * 3 * mint(2).pow(max(0, N - 3));
        }

        else ans *= 2 * mint(2).pow(max(N - 2, 0));
    }

    else {
        if (c[0] == 0 || c[N - 1] == 0) ans *= 2 * mint(2).pow(max(N - 1, 0));
        else ans *= 2 * mint(2).pow(N - 2);
    }

    cout << ans.val() << endl;
}