#include #include using namespace std; using namespace atcoder; // デバッグ表示 #define dump(x) cout << #x << ":" << (x) << endl; // 型定義 typedef long long ll; typedef pair P; // forループ #define REP(i,n) for(ll i=0; i<(ll)(n); ++i) // 定数宣言 const int INF = 1e9; const int MOD = 1e9+7; const ll LINF = 1e18; // modint using mint = modint1000000007; // using mint = modint998244353; // グラフ表現 using Graph = vector>; // グラフの辺表現 using Edge = map,int>; // n次元配列の初期化。第2引数の型のサイズごとに初期化していく。 template void Fill(A (&array)[N], const T &val){ std::fill( (T*)array, (T*)(array+N), val ); } // コンビネーションを計算する関数 ll pow(ll N, ll k) { ll res = 1; for (ll i = 0; i < k; ++i) res *= N; return res; } // 最大公約数 ll gcd(ll a,ll b){ if (a%b == 0) return(b); else return(gcd(b, a%b)); } // 最小公倍数 ll lcm(ll a, ll b){ return a/gcd(a, b) * b; } ll B[400][400]; int main() { cout << fixed << setprecision(15); ll K; cin >> K; ll N = 20; Fill(B, -1); vector C(N+1); C[0] = 1; B[200][200] = 0; for(ll t=1; t<=N; t++){ REP(i, 400){ REP(j, 400){ if(B[i][j] == t-1){ if(B[i-3][j-2] == -1) B[i-3][j-2] = t; if(B[i-3][j] == -1) B[i-3][j] = t; if(B[i-3][j+2] == -1) B[i-3][j+2] = t; if(B[i+3][j-2] == -1) B[i+3][j-2] = t; if(B[i+3][j] == -1) B[i+3][j] = t; if(B[i+3][j+2] == -1) B[i+3][j+2] = t; if(B[i-2][j-3] == -1) B[i-2][j-3] = t; if(B[i][j-3] == -1) B[i][j-3] = t; if(B[i+2][j-3] == -1) B[i+2][j-3] = t; if(B[i-2][j+3] == -1) B[i-2][j+3] = t; if(B[i][j+3] == -1) B[i][j+3] = t; if(B[i+2][j+3] == -1) B[i+2][j+3] = t; } } } ll count = 0; REP(i, 400){ REP(j, 400){ if(B[i][j] == t) count++; } } C[t] = count; } vector ans(20); // cout << 0 << " " << C[0] << endl; ans[0] = C[0]; ans[1] = C[1]; // cout << 0 << " " << ans[0] << endl; // cout << 1 << " " << ans[1] << endl; for(ll i=2; i<20; i++){ ans[i] = ans[i-2] + C[i]; // cout << i << " " << ans[i] << endl; } if(K < 10){ cout << ans[K] << endl; } else{ // 偶数のとき if(K%2 == 0){ ll u = (K-10)/2 + 1; mint ans2 = mint(17) * mint(10+K) * mint(u) / mint(2) - mint(14)*mint(u); ans2 *= mint(4); ans2 += mint(ans[8]); cout << ans2.val() << endl; } else{ ll u = (K-10)/2 + 1; mint ans2 = mint(17) * mint(11+K) * mint(u) / mint(2) - mint(14)*mint(u); ans2 *= mint(4); ans2 += mint(ans[9]); cout << ans2.val() << endl; } } return 0; }