B. Same Parity Summands

B. Same Parity Summands
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given two positive integers n (1≤n≤109) and k (1≤k≤100). Represent the number n as the sum of k positive integers of the same parity (have the same remainder when divided by 2).

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

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

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

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

Output
For each test case print:

YES and the required values ai, 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
inputCopy
8
10 3
100 4
8 7
97 2
8 8
3 10
5 3
1000000000 9
outputCopy
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

#include<iostream>
using namespace std;
int main()
{
	int n,a,b;
	cin>>n;
	while(n--)
	{
		cin>>a>>b;
		if(a%2==0)
		{
			if(b>a/2)
			{
				if(b%2==1)
				cout<<"NO"<<endl;
				else if(b<=a)
				{
					cout<<"YES"<<endl;
				    for(int i=0;i<b-1;i++)
			        cout<<"1"<<" ";
		        	cout<<a-(b-1)<<endl;
		     	}
		     	else
		     	{
		     		cout<<"NO"<<endl;
				 }
			}
			else
			{
			cout<<"YES"<<endl;
			for(int i=0;i<b-1;i++)
			cout<<"2"<<" ";
			cout<<a-2*(b-1)<<endl;
		    }
		}
		else
		{
			if(b%2==0)
			{
				cout<<"NO"<<endl;
			}
			else
			{
				if(b>a)
				cout<<"NO"<<endl;
				else
				{
				cout<<"YES"<<endl;
			    for(int i=0;i<b-1;i++)
			    cout<<"1"<<" ";
			    cout<<a-(b-1)<<endl;
			    }
			}
		}
	}
}