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>
using namespace std;
const int mod=998244353;
const int maxn=100005;
struct Matrix
{
    long long a[2][2];
};
Matrix mull(Matrix x,Matrix y)
{
    Matrix ans;
    for(int i=0;i<2;i++)
    {
        for(int j=0;j<2;j++)
        {
            ans.a[i][j]=0;
            for(int k=0;k<2;k++)
            {
                ans.a[i][j]=(ans.a[i][j]+x.a[i][k]*y.a[k][j])%mod;
            }
        }
    }
    return ans;
}
long long fib(long long m)
{
    if(m<1) return 0;
    if(m<3) return 1;
    Matrix ans;
    m--;
    for(int i=0;i<2;i++)
    {
        for(int j=0;j<2;j++)
            ans.a[i][j]=(i==j);
    }
    Matrix shit;
    shit.a[0][1]=shit.a[0][0]=shit.a[1][0]=1;
    shit.a[1][1]=0;
    while(m)
    {
        if(m&1) ans=mull(ans,shit);
        shit=mull(shit,shit);m=m>>1;
    }
    return (ans.a[1][0]+ans.a[1][1])%mod;
}
long long fast(long long a,long long num)
{
    long long ans=1;
    while(num){
        if(num&1) ans=ans*a%mod;
        a=a*a%mod;
        num=num>>1;
    }
    return ans;
}
int main()
{
    int n;
    long long m;
    scanf("%d%lld",&n,&m);
    int k=n-2;
    long long tmp=1;
    long long sum=fib(2*k+2);
    for(int i=1;i<=k;i++)
    {
        tmp=(m+i-1)%mod*tmp%mod*fast(i,mod-2)%mod;
        sum=sum+tmp*fib(2*k+2-2*i)%mod;
        sum=sum%mod;
    }
    sum=fib(m+2*k+2)-sum;
    sum=(sum%mod+mod)%mod;
    printf("%lld\n",sum);
    return 0;
}
