#include <map>
#include <set>
#include <array>
#include <cmath>
#include <queue>
#include <stack>
#include <tuple>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#include <functional>
#include <unordered_map>
using namespace std;

#define db double
#define il inline
#define fir first
#define sec second
#define eps (1e-10)
#define pb push_back
#define ll long long
#define mkp make_pair
#define eb emplace_back
#define pii pair<int, int>
#define lowbit(a) (a & (-a))
#define SZ(a) ((int)a.size())
#define ull unsigned long long
#define all(a) a.begin(), a.end()
#define split cout << "=========\n";
#define GG { cout << "NO\n"; return; }
#define pll pair<long long, long long>
#define equals(a, b) (fabs((a) - (b)) < eps)

constexpr int ON = 0;
constexpr int CW = -1;
constexpr int CCW = 1;
constexpr int BACK = 2;
constexpr int FRONT = -2;
const db pi = acos(-1.000);
constexpr int maxn = 5e5 + 50;
constexpr int INF = 0x3f3f3f3f;
constexpr ll LINF =  0x3f3f3f3f3f3f3f3f;
constexpr int mod = 998244353; /* 1e9 + 7 */
constexpr int dir[8][2] = {-1, 0, -1, 1, 0, 1, 1, 1, 1, 0, 1, -1, 0, -1, -1, -1};



namespace Linear_seq {
#define N maxn
typedef long long LL;
LL mod = 998244353;
#define rep(i, a, n) for (int i = a; i < n; i++)
#define pb push_back
typedef vector<int> VI;
typedef pair<int, int> PII;
// const int N = 10010;
LL res[N], base[N], _c[N], _md[N];
vector<int> Md;
LL powmod(LL a, LL b) {
  LL res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1)
      res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
void mul(LL *a, LL *b, int k) {
  rep(i, 0, k + k) _c[i] = 0;
  rep(i, 0, k) if (a[i]) rep(j, 0, k) _c[i + j] =
      (_c[i + j] + a[i] * b[j]) % mod;
  for (int i = k + k - 1; i >= k; i--)
    if (_c[i])
      rep(j, 0, SZ(Md)) _c[i - k + Md[j]] =
          (_c[i - k + Md[j]] - _c[i] * _md[Md[j]]) % mod;
  rep(i, 0, k) a[i] = _c[i];
}
int solve(LL n, VI a, VI b) {
  LL ans = 0, pnt = 0;
  int k = SZ(a);
  rep(i, 0, k) _md[k - 1 - i] = -a[i];
  _md[k] = 1;
  Md.clear();
  rep(i, 0, k) if (_md[i] != 0) Md.push_back(i);
  rep(i, 0, k) res[i] = base[i] = 0;
  res[0] = 1;
  while ((1LL << pnt) <= n)
    pnt++;
  for (int p = pnt; p >= 0; p--) {
    mul(res, res, k);
    if ((n >> p) & 1) {
      for (int i = k - 1; i >= 0; i--)
        res[i + 1] = res[i];
      res[0] = 0;
      rep(j, 0, SZ(Md)) res[Md[j]] = (res[Md[j]] - res[k] * _md[Md[j]]) % mod;
    }
  }
  rep(i, 0, k) ans = (ans + res[i] * b[i]) % mod;
  if (ans < 0)
    ans += mod;
  return ans;
}
VI BM(VI s) {
  VI C(1, 1), B(1, 1);
  int L = 0, m = 1, b = 1;
  rep(n, 0, SZ(s)) {
    LL d = 0;
    rep(i, 0, L + 1) d = (d + (LL)C[i] * s[n - i]) % mod;
    if (d == 0)
      ++m;
    else if (2 * L <= n) {
      VI T = C;
      LL c = mod - d * powmod(b, mod - 2) % mod;
      while (SZ(C) < SZ(B) + m)
        C.pb(0);
      rep(i, 0, SZ(B)) C[i + m] = (C[i + m] + c * B[i]) % mod;
      L = n + 1 - L;
      B = T;
      b = d;
      m = 1;
    } else {
      LL c = mod - d * powmod(b, mod - 2) % mod;
      while (SZ(C) < SZ(B) + m)
        C.pb(0);
      rep(i, 0, SZ(B)) C[i + m] = (C[i + m] + c * B[i]) % mod;
      ++m;
    }
  }
  return C;
}
int gao(VI a, LL n) {
  VI c = BM(a);
  c.erase(c.begin());
  rep(i, 0, SZ(c)) c[i] = (mod - c[i]) % mod;
  return solve(n, c, VI(a.begin(), a.begin() + SZ(c)));
}
}; // namespace Linear_seq

int n;
int ans;

void solve(int cas) {
    ll n; cin >> n;
    vector<int> a;
    a.pb(45);
    a.pb(4455);
    a.pb(407430);
    a.pb(36934785);
    a.pb(350210541);
    a.pb(427185456);
    a.pb(991901155);
    a.pb(110899952);
    a.pb(89411672);
    a.pb(401287887);
    a.pb(811189931);
    a.pb(286603985);
    a.pb(426843378);
    a.pb(418261776);
    a.pb(744223671);
    a.pb(302153364);
    a.pb(34240762);

// 689891595
// 101012015
// 748341487
    cout << Linear_seq::gao(a, n - 1);
}




signed main() {
    // cout << fixed << setprecision(10);
    ios::sync_with_stdio(false); cin.tie(nullptr);
    // int T; cin>>T; while (T--)
    solve(1);
    return 0;
}