There is a chip on the coordinate line. Initially, the chip is located at the point 00. You can perform any number of moves; each move increases the coordinate of the chip by some positive integer (which is called the length of the move). The length of the first move you make should be divisible by kk, the length of the second move — by k+1k+1, the third — by k+2k+2, and so on.

For example, if k=2k=2, then the sequence of moves may look like this: 0→4→7→19→440→4→7→19→44, because 4−0=44−0=4 is divisible by 2=k2=k, 7−4=37−4=3 is divisible by 3=k+13=k+1, 19−7=1219−7=12 is divisible by 4=k+24=k+2, 44−19=2544−19=25 is divisible by 5=k+35=k+3.

You are given two positive integers nn and kk. Your task is to count the number of ways to reach the point xx, starting from 00, for every x∈[1,n]x∈[1,n]. The number of ways can be very large, so print it modulo 998244353998244353. Two ways are considered different if they differ as sets of visited positions.

Input

The first (and only) line of the input contains two integers nn and kk (1≤k≤n≤2⋅1051≤k≤n≤2⋅105).

Output

Print nn integers — the number of ways to reach the point xx, starting from 00, for every x∈[1,n]x∈[1,n], taken modulo 998244353998244353.

Examples

input

Copy

8 1

output

Copy

1 1 2 2 3 4 5 6 

input

Copy

10 2

output

Copy

0 1 0 1 1 1 1 2 2 2 

Note

Let's look at the first example:

Ways to reach the point 11: [0,1][0,1];

Ways to reach the point 22: [0,2][0,2];

Ways to reach the point 33: [0,1,3][0,1,3], [0,3][0,3];

Ways to reach the point 44: [0,2,4][0,2,4], [0,4][0,4];

Ways to reach the point 55: [0,1,5][0,1,5], [0,3,5][0,3,5], [0,5][0,5];

Ways to reach the point 66: [0,1,3,6][0,1,3,6], [0,2,6][0,2,6], [0,4,6][0,4,6], [0,6][0,6];

Ways to reach the point 77: [0,2,4,7][0,2,4,7], [0,1,7][0,1,7], [0,3,7][0,3,7], [0,5,7][0,5,7], [0,7][0,7];

Ways to reach the point 88: [0,3,5,8][0,3,5,8], [0,1,5,8][0,1,5,8], [0,2,8][0,2,8], [0,4,8][0,4,8], [0,6,8][0,6,8], [0,8][0,8].

思路：

1，没什么规律，k+2在k+1的基础上转移来的，有点类似dp，所以就模拟一下k加的过程

2，b，c数组交替运用，详细的请看代码

代码：

#include
using namespace std;
#define int long long
#define vec vector
#define pi 3.1415926
#define rep(i,l,r) for(int i=l;i<=r;++i)
#pragma GCC optimize(2)//
#pragma GCC optimize(3,"Ofast","inline")//
const int maxj=2e5+100,mod=998244353,inf=0x7f7f7f7f7f7f7f;
int a[maxj],b[maxj],c[maxj];
void solve(){
    int n,k;cin>>n>>k;
    for(int i=k;i<=n;i+=k)a[i]++,b[i]++;//初始化数组
    int t=k*2+1;
    k++;
    bool ff=0;
    while(t<=n){//k+2在k+1基础上来的，类似于dp，有状态和转移方程之类的
        if(!ff){
            for(int i=t;i<=n;++i){
                c[i]=(b[i-k]+c[i-k])%mod;//b表示之前的状态，c表示当前k状态下，加多个k
                a[i]=(a[i]+c[i])%mod;//a进行累加
            }
            memset(b,0,sizeof(b));//清空b，防止对后续造成影响
        }else{
            for(int i=t;i<=n;++i){//c表示之前的状态，b表示当前k状态下，加多个k
                b[i]=(b[i-k]+c[i-k])%mod;
                a[i]=(a[i]+b[i])%mod;
            }
            memset(c,0,sizeof(c));
        }
        k++;
        ff=1-ff;
        t+=k;
    }
    for(int i=1;i<=n;++i)cout<' ';
    cout<<'\n';
}
signed main(){
    ios::sync_with_stdio(0);//能不用c语言的输入输出，尽量不要用
    cin.tie(0);
    cout.tie(0);
    int t;
    t=1;
    // cin>>t;
    while(t--)solve();
    return 0;
}