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<iostream>
#include<sstream>
#include<fstream>
#include<string>
#include<vector>
#include<deque>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<algorithm>
#include<functional>
#include<utility>
#include<bitset>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<cstdio>

using namespace std;

#define REP(i,n) for(int i=0;i<int(n);i++)
#define foreach(c,itr) for(__typeof((c).begin()) itr=(c).begin();itr!=(c).end();itr++)
typedef long long ll;
typedef pair<int,int> P;

int d[3010][3010],r[3010];

int main(void){
	int i,j;

	cin.tie(0);
	ios_base::sync_with_stdio(false);

	ll n,m;
	cin >> n >> m;

	ll M=998244353;
	d[1][1]=1; d[1][2]=1;
	for(i=3;i<=m;i++) d[1][i]=(d[1][i-1]+d[1][i-2])%M;

	for(i=2;i<=n;i++){
		r[1]=d[i-1][1]%M;
		for(j=2;j<=m;j++) r[j]=(r[j]+r[j-1])%M;

		for(j=1;j<m;j++){
			d[i][j]=r[j];
		}
	}


	cout << d[n][m] << endl;


	return 0;
}