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++ (GCC 5.4.1) (judge verdict: Runtime Error)

#include <queue>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define LL long long
const int MAXN=1e5+5;
const int mod=998244353;
LL fac[100005],inv[100005],f[5*MAXN];
LL qkpow(LL x, LL power, LL mod) {
    x %= mod;
    LL ans = 1;
    for (; power; power >>= 1, (x *= x) %= mod)
        if (power & 1)
            (ans *= x) %= mod;
    return ans;
}
void init() {
    fac[0] = 1;
    for (int i = 1; i < MAXN; i++) fac[i] = fac[i - 1] * i % mod;
    inv[0] = 1;
    for (int i = 1; i < MAXN; i++) inv[i] = qkpow(fac[i], mod - 2, mod);
}
LL C(LL n, LL m) {
    if (n < m)
        return 0;
    return fac[n] * inv[m] % mod * inv[n - m] % mod;
}
LL n,m,ans;
int main() {
	scanf("%lld %lld",&n,&m);
	f[1]=1,f[2]=1;
	init();
	for(int i=3;i<=m;i++)f[i]=f[i-1]+f[i-2],f[i]%=mod;
	if(n==1){
		printf("%lld\n",f[m]);
		return 0;
	}
	for(int i=0;i<m;i++){
		ans=(ans+C(n+i-2,n-2)*f[m-i])%mod;
	}
	printf("%lld\n",ans);
	return 0;
}