#include <iostream>
#include <cstdio>
#include <string>
#include <algorithm>
#include <utility>
#include <cmath>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <tuple>
#include <numeric>
#include <functional>
using namespace std;
typedef long long ll;
typedef vector<ll> vl;
typedef vector<vector<ll>> vvl;
typedef pair<ll, ll> P;
#define rep(i, n) for(ll i = 0; i < n; i++)
#define exrep(i, a, b) for(ll i = a; i <= b; i++)
#define out(x) cout << x << endl
#define exout(x) printf("%.10f\n", x)
#define chmax(x, y) x = max(x, y)
#define chmin(x, y) x = min(x, y)
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define pb push_back
#define re0 return 0
const ll mod = 998244353;
const ll INF = 1e16;

// a^n (mod.MOD)を求める。計算量はO(logn)
ll modpow(ll a, ll n, ll MOD = mod) {
    if(n == 0) {
        return 1;
    }
    if(n % 2 == 1) {
        return a * modpow(a, n-1, MOD) % MOD;
    }
    return modpow(a, n/2, MOD) * modpow(a, n/2, MOD) % MOD;
}

int main() {
    string s;
    cin >> s;
    ll n = s.size();

    // f(i,j):下からi桁目の数s[n-i]が、下からj桁目(1 <= j <= i)として使われる回数
    // f(i, j) = 2^(n-i)*Comb(i-1, j-1)

    // 求める値は、
    // Σs[n-i]*2^(n-i)*(ΣComb(i-1, j-1)*10^(j-1) (1 <= j <= i)) (1 <= i <= n)

    //   ΣComb(i-1, j-1)*10^(j-1) (1 <= j <= i) 
    // = ΣComb(i-1, j)*10^j (0 <= j <= i-1)
    // = (10 + 1)^(i-1) = 11^(i-1)

    // ∴ Σs[n-i]*2^(n-i)*11^(i-1) (1 <= i <= n)

    ll ans = 0;
    exrep(i, 1, n) {
        ans += (s[n - i] - '0') * modpow(2LL, n - i) % mod * modpow(11LL, i-1);
        ans %= mod;
    }

    out(ans);
    re0;
}