# CF1106F Lunar New Year and a Recursive Sequence 线性递推 + k次剩余

$f_k = 3^s(mod\;998244353)$$m = 3^t$，那么有$3^{sp} = 3^t (mod\;998244353)$

#include <map>
#include <queue>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

#define de double
#define ll long long
#define ri register int
#define rep(io, st, ed) for(ri io = st; io <= ed; io ++)
#define drep(io, ed, st) for(ri io = ed; io >= st; io --)

#define gc getchar
int p = 0, w = 1; char c = gc();
while(c > '9' || c < '0') { if(c == '-') w = -1; c = gc(); }
while(c >= '0' && c <= '9') p = p * 10 + c - '0', c = gc();
return p * w;
}

const int sid = 105;
const int mod = 998244353;
const int g = 3;

inline int mul(int a, int b) { return 1ll * a * b % mod; }
inline int fp(int a, int k) {
int ret = 1;
for( ; k; k >>= 1, a = mul(a, a))
if(k & 1) ret = mul(ret, a);
return ret;
}

int n, m, k, b[sid];
struct mtx {
int f[sid][sid];
mtx() {}
mtx(int flag) {
if(flag == 0) {
for(int i = 0; i < k; i ++)
for(int j = 0; j < k; j ++)
f[i][j] = 0;
}
if(flag == 1) {
for(int i = 0; i < k; i ++)
for(int j = 0; j < k; j ++)
f[i][j] = (i == j) ? 1 : 0;
}
}
int* operator [] (const int x) {
return f[x];
}
friend mtx operator * (mtx a, mtx b) {
mtx c(0);
for(int i = 0; i < k; i ++)
for(int j = 0; j < k; j ++)
for(int p = 0; p < k; p ++)
c[i][j] = (c[i][j] + 1ll * a[i][p] * b[p][j] % (mod - 1)) % (mod - 1);
return c;
}
} A, B;

inline mtx fp(mtx a, int k) {
mtx ret(1);
for( ; k; k >>= 1, a = a * a)
if(k & 1) ret = ret * a;
return ret;
}

map <int, int> H;
inline int BSGS(int A, int B) {
H.clear(); H[1] = 0;
int Ai = 1, Aj = 1, m = ceil(sqrt(mod));
for(ri i = 1; i < m; i ++) Ai = 1ll * Ai * A % mod, H[1ll * Ai * B % mod] = i;
Ai = 1ll * Ai * A % mod;
for(ri i = 1; i <= m; i ++) {
Aj = 1ll * Aj * Ai % mod;
if(H[Aj]) return 1ll * i * m - H[Aj];
}
}

inline int gcd(int a, int b) {
return b ? gcd(b, a % b) : a;
}

inline void exgcd(int &x, int &y, int a, int b) {
if(!b) { x = 1; y = 0; return; }
exgcd(y, x, b, a % b); y -= a / b * x;
}

inline void Solve() {
A[0][k - 1] = 1;
rep (i, 0, k - 1) B[i][k - 1] = b[k - i];
rep (i, 0, k - 2) B[i + 1][i] = 1;

A = A * fp(B, n - k);

int p = A[0][k - 1], t = BSGS(g, m);
if(t % gcd(p, mod - 1)) printf("-1\n");
else {
int gd = gcd(p, mod - 1);
int x, y, a = p, b = mod - 1;
t /= gd; a /= gd; b /= gd;
exgcd(x, y, a, b);
x = (x + mod - 1) % (mod - 1);
x = 1ll * x * t % (mod - 1);
printf("%d\n", fp(g, x));
}
}

int main() {