#include <bits/stdc++.h>

using namespace std;
using ll = long long;

const ll modc = 998244353;
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) {
        ll t;
        ip >> t;
        m = mint(t);
        return ip;
    }
    ll val(){
        return x;
    }
};

int main(){

    int N;
    string S;
    mint ans=0;
    cin >> N >> S;
    vector<vector<mint>> dp(N+1, vector<mint>(3));

    for (int i=1; i<=N; i++){
        char c = S[i-1];
        if (c != '?'){
            dp[i][(c-'0')%3] += 1;
            for (int j=0; j<3; j++){
                dp[i][(c-'0'+j)%3] += dp[i-1][j];
            }
        }
        else{
            for (int j=0; j<10; j++){
                dp[i][j%3] += 1;
                for (int k=0; k<3; k++){
                    dp[i][(j+k)%3] += dp[i-1][k];
                }
            }
        }
        ans += dp[i][0];
    }

    cout << ans << endl;

    return 0;
}