#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 = 200000;
    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];
    }
    int T;
    cin >> T;
    while(T--){
        ll n;
        cin >> n;
        // 桁和の最低値
        ll mn = n + 1;
        string str = '0' + to_string(mn);
        // mn以上の値で桁DP
        int len = str.size(), tot = len * 9;
        vector dp(len + 1, vector(2, vector(10, vector<ll>(tot + 1, 1ll << 62))));
        dp[0][1][0][0] = 0;
        for(int i = 0; i < len; i++){
            int th = 9 * i;
            for(int j = 0; j < 2; j++){
                for(int k = 0; k < 10; k++){
                    int low = j ? max(k, str[i] - '0') : k;
                    int low2 = str[i] - '0';
                    for(int l = 0; l <= th; l++){
                        if(dp[i][j][k][l] >> 62) continue;
                        for(int nv = low; nv < 10; nv++){
                            dp[i + 1][j & (low2 == nv)][nv][l + nv] 
                            = min(dp[i + 1][j & (low2 == nv)][nv][l + nv], dp[i][j][k][l] * 10 + nv);
                        }
                    }
                }
            }
        }
        mn = 1ll << 62;
        for(int i = 0; i < 2; i++){
            for(int j = 0; j < 10; j++){
                for(int k = 0; k <= tot; k++){
                    if(tb[k]) mn = min(mn, dp[len][i][j][k]);
                }
            }
        }
        ll keta = n + 1 + max(0ll, (mn - 9 * (n + 1) + 8) / 9);
        mint ans;
        ans += (mint(10).pow(keta) - 1) / 9;
        mn -= keta;
        ll keta2 = mn / 8;
        ans += (mint(10).pow(keta2) - 1) * 8 / 9;
        mn -= keta2 * 8;
        ans += mint(10).pow(keta2) * mn;
        cout << ans.val() << '\n';
    }
}