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

#include <bits/stdc++.h>
#include <atcoder/all>
using namespace std;
using namespace atcoder;
using ll = long long;
using pll = pair<ll, ll>;
using pii = pair<int, int>;
using mint = modint998244353;
constexpr ll mod = 998244353;
using MINT = modint1000000007;
constexpr ll MOD = 1000000007;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};

template <typename T> void print(vector<T> A);

mint solve(string S) {
    int N = S.size();
    if (S[0]=='?') {
        string T1 = S, T2 = S;
        T1[0] = 'o';
        T2[0] = 'x';
        return solve(T1)+solve(T2);
    }

    

    mint dp[1000002][2][2][2];

    if (S[0] == 'o') {
        dp[0][1][1][0]=1;
    } else {
        dp[0][0][1][0]=1;
    } 
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < 2; j++) {
            for (int k = 0; k < 2; k++) {
                for (int l = 0; l < 2; l++) {
                    if (S[i] != 'x') { //oか?のとき
                        if (j == 1) {
                            dp[i+1][1][1][l]+=dp[i][j][k][l];
                        } else {
                            if (k == 1) {
                                dp[i+1][1][0][l]+=dp[i][j][k][l];
                            } else {
                                if (l<1)dp[i+1][1][0][l+1]+=dp[i][j][k][l];
                            }
                        }
                    }

                    if (S[i] != 'o') { //xか?のとき
                        if (j == 0) {
                            dp[i+1][j][1][l]+=dp[i][j][k][l];
                        } else {
                            if (k == 1) {
                                dp[i+1][0][0][l]+=dp[i][j][k][l];
                            } else {
                                if (l<1)dp[i+1][0][0][l+1]+=dp[i][j][k][l];
                            }
                        }
                    }
                }
            }
        }
    }

    mint ret = 0;
    for (int j = 0; j < 2; j++) {
        for (int k = 0; k < 2; k++) {
            for (int l = 0; l < 2; l++) {
                //cout<<j<<' '<<k<<' '<<l<<' '<<dp[N][j][k][l].val()<<endl;
                ret+=dp[N][j][k][l];
            }
        }
    }
    return ret;


}

int main()
{
    ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    int N;
    cin >> N;
    string S;
    cin >> S;

    cout<<solve(S).val()<<endl;
    

    
}

template <typename T>
void print(vector<T> A)
{
    for (int i = 0; i < A.size() - 1; i++)
    {
        cout << A[i] << ' ';
    }
    cout << A[A.size() - 1] << endl;
    return;
}