#include using namespace std; #define double long double using ll = long long; using VB = vector; using VVB = vector; using VC = vector; using VI = vector; using VVI = vector; using VVVI = vector; using VL = vector; using VVL = vector; using VVVL = vector; using VVVVL = vector; using VD = vector; using VVD = vector; using VVVD = vector; using P = pair; #define REP(i, n) for (int i = 0; i < (int)(n); i++) #define FOR(i, a, b) for (ll i = a; i < (ll)(b); i++) #define ALL(a) (a).begin(),(a).end() constexpr int INF = 1001001001; constexpr ll LINF = 1001001001001001001ll; constexpr int DX[] = {1, 0, -1, 0}; constexpr int DY[] = {0, 1, 0, -1}; template< typename T1, typename T2> inline bool chmax(T1 &a, T2 b) {return a < b && (a = b, true); } template< typename T1, typename T2> inline bool chmin(T1 &a, T2 b) {return a > b && (a = b, true); } const ll MOD = 998244353; const int MAX_N = 200010; int par[MAX_N]; int rnk[MAX_N]; int siz[MAX_N]; void init(int n) { REP(i,n) { par[i] = i; rnk[i] = 0; siz[i] = 1; } } int find(int x) { if (par[x] == x) { return x; } else { return par[x] = find(par[x]); } } void unite(int x, int y) { x = find(x); y = find(y); if (x == y) return; int s = siz[x] + siz[y]; if (rnk[x] < rnk[y]) { par[x] = y; } else { par[y] = x; if (rnk[x] == rnk[y]) rnk[x]++; } siz[find(x)] = s; } bool same(int x, int y) { return find(x) == find(y); } int size(int x) { return siz[find(x)]; } int main() { ios::sync_with_stdio(false); std::cin.tie(nullptr); int n; cin >> n; // dp[i][j][k] aiまでで長さがjで和がkの場合の数 VVVL dp(2, VVL(500, VL(500, 0))); dp[0][0][0] = 1; REP(i, n) { VVL d(500, VL(500, 0)); REP(j, 500) { REP(k, 500) { if (j + 2 < 500 && k + 1 < 500) d[j + 2][k + 1] = (d[j + 2][k + 1] + dp[i % 2][j][k]) % MOD; if (j + 2 < 500 && k + 10 < 500) d[j + 2][k + 10] = (d[j + 2][k + 10] - dp[i % 2][j][k] + MOD) % MOD; if (j + 3 < 500 && k + 10 < 500) d[j + 3][k + 10] = (d[j + 3][k + 10] + dp[i % 2][j][k]) % MOD; if (j + 3 < 500 && k + 100 < 500) d[j + 3][k + 100] = (d[j + 3][k + 100] - dp[i % 2][j][k] + MOD) % MOD; if (j + 4 < 500 && k + 100 < 500) d[j + 4][k + 100] = (d[j + 4][k + 100] + dp[i % 2][j][k]) % MOD; } } REP(j, 500) { dp[(i + 1) % 2][j][0] = 0; REP(k, 499) { dp[(i + 1) % 2][j][k + 1] = (dp[(i + 1) % 2][j][k] + d[j][k + 1]) % MOD; } } /* REP(j, 10) { REP(k, 20) cout << dp[(i + 1) % 2][j][k] << " "; cout << endl; } cout << endl; */ } ll ans = 0; REP(i, 499) ans = (ans + dp[n % 2][i + 1][i]) % MOD; cout << ans << endl; }