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.

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.

Input

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



> $n \quad m$

Examples

Input

4 7


Output

176


Input

12 20


Output

174174144


Input

16 30


Output

102292850
#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;
}