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>
using namespace std;
typedef signed long long ll;

#undef _P
#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
//-------------------------------------------------------

ll N,M;
ll mo=998244353; //2^23*7*17+1
ll D[202020];

ll modpow(ll a, ll n) {
	ll r=1;
	while(n) r=r*((n%2)?a:1)%mo,a=a*a%mo,n>>=1;
	return r;
}

ll comb(int P_,int Q_) {
	if(P_<0 || Q_<0 || Q_>P_) return 0;
	ll ret=1;
	Q_=min(Q_,P_-Q_);
	for(int i=1;i<=Q_;i++) ret=ret*(P_+1-i)%mo*modpow(i,mo-2)%mo;
	return ret;
}

ll fib(ll k,ll a1=1,ll a2=1) {
	if(k==1) return a1;
	if(k==2) return a2;
	k-=2;
	ll R[2][2]={{1,0},{0,1}};
	ll A[2][2]={{1,1},{1,0}};
	
	while(k) {
		if(k%2) {
			ll S[2][2];
			S[0][0]=(A[0][0]*R[0][0]+A[0][1]*R[1][0])%mo;
			S[0][1]=(A[0][0]*R[0][1]+A[0][1]*R[1][1])%mo;
			S[1][0]=(A[1][0]*R[0][0]+A[1][1]*R[1][0])%mo;
			S[1][1]=(A[1][0]*R[0][1]+A[1][1]*R[1][1])%mo;
			swap(S,R);
		}
		ll T[2][2];
		T[0][0]=(A[0][0]*A[0][0]+A[0][1]*A[1][0])%mo;
		T[0][1]=(A[0][0]*A[0][1]+A[0][1]*A[1][1])%mo;
		T[1][0]=(A[1][0]*A[0][0]+A[1][1]*A[1][0])%mo;
		T[1][1]=(A[1][0]*A[0][1]+A[1][1]*A[1][1])%mo;
		swap(T,A);
		k>>=1;
	}
	return (R[0][0]+R[0][1])%mo;
	
	
}

void solve() {
	int i,j,k,l,r,x,y; string s;
	
	cin>>N>>M;
	
	M+=2*(N-1);
	ll F=fib(M);
	if(N>1000) return;
	for(i=1;i<N;i++) {
		M-=2;
		F=(F+mo-comb(M+i,i-1))%mo;
	}
	cout<<F<<endl;
	
	
}


int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false), cin.tie(0);
	FOR(i,argc-1) s+=argv[i+1],s+='\n';
	FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	solve(); return 0;
}