#include<iostream>
#include<cstdio>
using namespace std;
int n,k,ans[350005],cnt;
int main(){
    scanf("%d%d",&k,&n);
    if(k&1){
        cnt=n;
        for(int i=1;i<=n;i++)
            ans[i]=k/2+1;
        for(int i=1;i<=n/2;i++){
            if(ans[cnt]==1){
                cnt--;
                continue;
            }
            ans[cnt]--;
            for(int j=cnt+1;j<=n;j++)
                ans[j]=k;
            cnt=n;
        }
        for(int i=1;i<=cnt;i++) printf("%d ",ans[i]);
        return 0;
    }
    //tttteeeeeepppppaaaaaaaaannnnnnn
    printf("%d ",k/2);
    for(int i=2;i<=n;i++)
        printf("%d ",k);
    return 0;
}
### Prompt

Your challenge is to write a Cpp solution to the following problem:

In Finite Encyclopedia of Integer Sequences (FEIS), all integer sequences of lengths between 1 and N (inclusive) consisting of integers between 1 and K (inclusive) are listed.

Let the total number of sequences listed in FEIS be X. Among those sequences, find the (X/2)-th (rounded up to the nearest integer) lexicographically smallest one.

Constraints

* 1 \leq N,K \leq 3 × 10^5
* N and K are integers.

Input

Input is given from Standard Input in the following format:


K N


Output

Print the (X/2)-th (rounded up to the nearest integer) lexicographically smallest sequence listed in FEIS, with spaces in between, where X is the total number of sequences listed in FEIS.

Examples

Input

3 2


Output

2 1


Input

2 4


Output

1 2 2 2


Input

5 14


Output

3 3 3 3 3 3 3 3 3 3 3 3 2 2

### Response

```cpp
#include<iostream>
#include<cstdio>
using namespace std;
int n,k,ans[350005],cnt;
int main(){
    scanf("%d%d",&k,&n);
    if(k&1){
        cnt=n;
        for(int i=1;i<=n;i++)
            ans[i]=k/2+1;
        for(int i=1;i<=n/2;i++){
            if(ans[cnt]==1){
                cnt--;
                continue;
            }
            ans[cnt]--;
            for(int j=cnt+1;j<=n;j++)
                ans[j]=k;
            cnt=n;
        }
        for(int i=1;i<=cnt;i++) printf("%d ",ans[i]);
        return 0;
    }
    //tttteeeeeepppppaaaaaaaaannnnnnn
    printf("%d ",k/2);
    for(int i=2;i<=n;i++)
        printf("%d ",k);
    return 0;
}
```