B. Same Parity Summands

Description

You are given two positive integers n (1≤n≤10^9) 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

input

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

output

ES
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

Solution

思路：

因为结果都要同一类数，非奇即偶，所以对于有答案的每组结果形式进行固定，固定形式为都是奇数的话，除第一个数其他数全为1，同理偶数则为2.

则本题主要对n减去k-1个1或者2的奇偶进行判断。若减去k-1个1仍是奇数或减去k-1个2仍为偶数则有答案并输出答案结构。

代码：

#include <iostream>
#include <cstdio>
#define ll long long
using namespace std;
int main(){
    ll t,n,k;
    scanf("%lld",&t);
    while(t--){
        cin>>n>>k;
        if( (n-(k-1))&1 && n-k+1 > 0){
            printf("YES\n");
            printf("%lld ",n-k+1);
            while(--k) printf("1 ");
            printf("\n");
            continue;
        }
        if( !((n-k*2+2)&1) && n-k*2+2 > 0){
            printf("YES\n");
            printf("%lld ",n-2*k+2);
            while(--k) printf("2 ");
            printf("\n");
            continue;
        }
        printf("NO\n");
    }
    return 0;
}