Problem:
Max Score: $1400$ Points

Problem Statement

E869120 defined a sequence $a$ like this:

$a_1=a_2=1$, $a_{k+2}=a_{k+1}+a_k \ (k \ge 1)$

He also defined sequences $d_1, d_2, d_3, \dots , d_n$, as the following recurrence relation :

$d_{1, j} = a_j$

$d_{i, j} = \sum_{k = 1}^j d_{i - 1, k} \ (i \ge 2)$

You are given integers $n$ and $m$. Please calculate the value of $d_{n, m}$.

Since the answer can be large number, print the answer modulo $998,244,353$.

Can you solve this problem???

Input

The input is given from standard input in the following format.

$n \quad m$

Output

Print $d_{n, m}$ modulo $998,244,353$.

Constraints

$1 \le n \le 200,000$

$1 \le m \le 10^{18}$

Subtasks

Subtask 1 [ $100$ points ]

The testcase in this subtask satisfies $1 \le n, m \le 3,000$.

Subtask 2 [ $170$ points ]

The testcase in this subtask satisfies $1 \le m \le 200,000$.

Subtask 3 [ $230$ points ]

The testcase in this subtask satisfies $1 \le n \le 3$.

Subtask 4 [ $420$ points ]

The testcase in this subtask satisfies $1 \le n \le 1000$.

Subtask 5 [ $480$ points ]

There are no additional constraints.

Sample Input 1

4 7

Sample Output 1

176

The sequence is following:

$d_{i, 1}$

$d_{i, 2}$

$d_{i, 3}$

$d_{i, 4}$

$d_{i, 5}$

$d_{i, 6}$

$d_{i, 7}$

$d_1$

1

1

2

3

5

8

13

$d_2$

1

2

4

7

12

20

33

$d_3$

1

3

7

14

26

46

79

$d_4$

1

4

11

25

51

97

176

As a result, the answer is $d_{4, 7} = 176$.

Sample Input 2

12 20

Sample Output 2

174174144

Interesting number. It only contains $1$, $4$ and $7$.

Sample Input 3

16 30

Sample Output 3

102292850

You can calculate that $d_{16, 30} = 1,193,004,294,685$.

Please don't forget printing the answer mod $998,244,353$.

Writer：square1001

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define mod 998244353
#define N 4000050
using namespace std;
typedef long long LL;
inline LL rd() { LL r; cin >> r; return r; }
inline void inc(LL &x,LL y) { x = ( x + y ) % mod; }
inline void dec(LL &x,LL y) { x = ( x - y + mod ) % mod; }

LL n,m,C[N],f[N];

struct Matrix{ LL d[4][4]; }id,A,B;
Matrix operator* (Matrix p1, Matrix p2) {
	Matrix ret = id;
	for (int i=1;i<=2;i++)
		for (int j=1;j<=2;j++)
			for (int k=1;k<=2;k++) 
				inc( ret.d[i][j] , p1.d[i][k] * p2.d[k][j] % mod );
	return ret;
}

Matrix Mqp(Matrix a,LL b) {
	Matrix ret = id; ret.d[1][1] = ret.d[2][2] = 1LL;
	while (b) {
		if (b&1) ret = ret * a;
		b >>= 1, A = a * a;
	}
	return ret;
}

LL qp(LL a,LL b) {
	LL ret = 1LL;
	a %= mod;
	while (b) {
		if (b&1) ret = ret * a % mod;
		b >>= 1, a = a * a % mod;
	}
	return ret;
}

LL F(LL x) {
	if (x <= 2) return 1LL;
	A.d[1][1] = A.d[1][2] = 1LL;
	B.d[1][1] = B.d[2][1] = B.d[1][2] = 1LL;
	B = Mqp(B,x-2);
	A = A * B;
	return A.d[1][1];
}

int main() {
	n = rd(), m = rd();
	C[n-1] = 1LL;
	for (LL _=n-2;_>=1;_--) 
		C[_] = C[_+1] * ( (n+m-_-2) % mod ) % mod * qp(n-_-1,mod-2) % mod;
	
	f[1] = f[2] = 1LL;
	for (int _=3;_<=2*n;_++) f[_] = ( f[_-1] + f[_-2] ) % mod;
	
	LL ans = F(2*n+m-2);
	for (int k=1;k<=n-1;k++) 
		dec(ans, f[2*k] * C[k] % mod);
	printf("%d\n",(int)ans);
	return 0;
}