B. Same Parity Summands

B. Same Parity Summands

B. Same Parity Summands

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given two positive integers nn (1≤n≤1091≤n≤109) and kk (1≤k≤1001≤k≤100). Represent the number nn as the sum of kk positive integers of the same parity (have the same remainder when divided by 22).

In other words, find a1,a2,…,aka1,a2,…,ak such that all ai>0ai>0, n=a1+a2+…+akn=a1+a2+…+ak and either all aiai are even or all aiai are odd at the same time.

If such a representation does not exist, then report it.

Input

The first line contains an integer tt (1≤t≤10001≤t≤1000) — the number of test cases in the input. Next, tt test cases are given, one per line.

Each test case is two positive integers nn (1≤n≤1091≤n≤109) and kk (1≤k≤1001≤k≤100).

Output

For each test case print:

• YES and the required values aiai, if the answer exists (if there are several answers, print any of them);
• NO if the answer does not exist.

The letters in the words YES and NO can be printed in any case.

Example

input

Copy

8
10 3
100 4
8 7
97 2
8 8
3 10
5 3
1000000000 9

output

Copy

YES
4 2 4
YES
55 5 5 35
NO
NO
YES
1 1 1 1 1 1 1 1
NO
YES
3 1 1
YES
111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111110 111111120

=========================================================================

分别讨论奇数偶数情况，奇数的话，先都放上1，然后每个数字再分别填上2，保证奇偶性相同，这就要求剩下的数字必须是偶数，否则无法构造，输出的时候只需要把全部2都加在一个位置上即可。然后讨论偶数的情况，先都放上2，剩下的必须也是2的倍数。

#include<iostream>
#include<cstdio>
#include<cstring>
# include<iomanip>

#include<algorithm>
#define  mo 998244353;
using namespace std;

typedef long long int ll;


int main()
{


  int t;
  cin>>t;

  while(t--)
  {
      int n,k;

      cin>>n>>k;

      if(k>n)
      {
          cout<<"NO"<<endl;
      }
      else if(k==n)
      {
          cout<<"YES"<<endl;

          for(int i=1;i<=n;i++)
          {
              cout<<1<<" ";

          }

          cout<<endl;
      }
      else
      {

          int flag1=0,flag2=0;

          int temp1=n-k;

          if(temp1<0||temp1%2)
          flag1=1;

          int temp2=n-2*k;

          if(temp2<0||temp2%2)
            flag2=1;

          if(flag1&&flag2)
          {
              cout<<"NO"<<endl;

          }
          else
          {
              
              cout<<"YES"<<endl;
              
              if(!flag1)
              {
                  for(int i=1;i<k;i++)
                  {
                      cout<<1<<" ";
                  }
                  cout<<1+temp1<<endl;
              }
              else
              {


                  for(int i=1;i<k;i++)
                  {
                      cout<<2<<" ";
                  }


                  cout<<2+temp2<<endl;

              }
          }

      }
  }

    return 0;
}