#include <bits/stdc++.h>
#include <atcoder/all>
using namespace std;
using ll = long long;
using mint = atcoder::modint998244353;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    constexpr int r = 300;
    auto check = [&](int v){
        int mn = 9, res = 0;
        while(v){
            int r = v % 10;
            res += r;
            if(r > mn) return -1;
            mn = min(mn, r);
            v /= 10;
        }
        return res;
    };
    vector<bool> tb(r + 1);
    for(int i = 0; i <= 9; i++) tb[i] = true;
    for(int i = 10; i <= r; i++){
        int s = check(i);
        if(s == -1) continue;
        tb[i] = tb[s];
    }
    constexpr ll th = 1'000'000'000'000'000ll;
    vector<ll> S;
    auto dfs = [&](auto dfs, ll v, int sv, int low) -> void {
        if(tb[sv]) S.emplace_back(v);
        if(v >= th) return;
        for(int i = low; i <= 9; i++){
            dfs(dfs, v * 10 + i, sv + i, i);
        }
    };
    dfs(dfs, 0, 0, 1);
    sort(S.begin(), S.end());
    int T;
    cin >> T;
    while(T--){
        ll n;
        cin >> n;
        ll mn = *lower_bound(S.begin(), S.end(), n + 1);
        ll keta = max(n + 1, (mn + 8) / 9);
        mint ans;
        ans += (mint(10).pow(keta) - mint(1)) / 9;
        mn -= keta;
        ll keta2 = mn / 8;
        ans += (mint(10).pow(keta2) - mint(1)) * mint(8) / 9;
        mn -= keta2 * 8;
        ans += mint(10).pow(keta2) * mn;
        cout << ans.val() << '\n';
    }
}