#include using namespace std; using ll = long long; using ld = long double; // -------------------------------------------------------- template bool chmax(T& a, const T b) { if (a < b) { a = b; return 1; } return 0; } template bool chmin(T& a, const T b) { if (b < a) { a = b; return 1; } return 0; } #define FOR(i,l,r) for (ll i = (l); i < (r); ++i) #define RFOR(i,l,r) for (ll i = (r)-1; (l) <= i; --i) #define REP(i,n) FOR(i,0,n) #define RREP(i,n) RFOR(i,0,n) #define ALL(c) (c).begin(), (c).end() #define RALL(c) (c).rbegin(), (c).rend() #define SORT(c) sort(ALL(c)) #define RSORT(c) sort(RALL(c)) #define MIN(c) *min_element(ALL(c)) #define MAX(c) *max_element(ALL(c)) #define SUMLL(c) accumulate(ALL(c), 0LL) #define COUNT(c,v) count(ALL(c),(v)) #define SZ(c) ((ll)(c).size()) #define BIT(b,i) (((b)>>(i)) & 1) #define PCNT(b) __builtin_popcountll(b) #define CIN(c) cin >> (c) #define COUT(c) cout << (c) << '\n' #define debug(x) cerr << "l." << __LINE__ << " : " << #x << " = " << (x) << '\n' ll llceil(ll a, ll b) { return (a + b - 1) / b; } ll bitlen(ll b) { if (b <= 0) { return 0; } return (64LL - __builtin_clzll(b)); } string toupper(const string& S) { string T(S); REP(i,SZ(T)) T[i] = toupper(T[i]); return T; } string tolower(const string& S) { string T(S); REP(i,SZ(T)) T[i] = tolower(T[i]); return T; } using P = pair; using VP = vector

; using VVP = vector; using VS = vector; using VVS = vector; using VLL = vector; using VVLL = vector; using VVVLL = vector; using VB = vector; using VVB = vector; using VVVB = vector; using VD = vector; using VVD = vector; using VVVD = vector; using VLD = vector; using VVLD = vector; using VVVLD = vector; static const double EPS = 1e-10; static const double PI = acos(-1.0); static const ll MOD = 1000000007; // static const ll MOD = 998244353; static const ll INF = (1LL << 62) - 1; // 4611686018427387904 - 1 // -------------------------------------------------------- #include using namespace atcoder; using mint = modint; using VM = vector; using VVM = vector; /** * @brief 行列累乗 * d x d の正方行列 A に対して A^n を O(k^3 log n) で求める * * @tparam T 行列要素の型 e.g.) mint, ll * @param A 正方行列 * @param n 指数 * @return vector> A^n の計算結果 */ template vector> mat_exp(vector> A, ll n) { using VT = vector; using VVT = vector; const ll d = (ll)A.size(); VVT B(d, VT(d, 0)); REP(i,d) B[i][i] = 1; // 単位行列で初期化 auto mat_mul = [&](const VVT& A, const VVT& B) -> VVT { VVT C(d, VT(d, 0)); REP(i,d) REP(k,d) REP(j,d) { C[i][j] += A[i][k] * B[k][j]; } return C; }; // e.g.) n = 11, B = A^(2^3) + A^(2^1) + A^(2^0) (11 = 2^3 + 2^1 + 2^0) while (n > 0) { if (n & 1) B = mat_mul(B, A); // 欲しいタイミングで拾う A = mat_mul(A, A); n >>= 1; } return B; }; int main() { ios::sync_with_stdio(false); cin.tie(nullptr); cout << fixed << setprecision(15); ll N, M; cin >> N >> M; mint::set_mod(M); VVM A = {{1, 1}, {1, 0}}; auto An = mat_exp(A, N-2); ll ans = An[0][0].val(); COUT(ans); return 0; }