Problem: It is the hard version of the problem. The only difference is that in this version 3 ≤ k ≤ n.

You are given a positive integer n. Find k positive integers a_1, a_2, …, a_k, such that:

  * a_1 + a_2 + … + a_k = n 
  * LCM(a_1, a_2, …, a_k) ≤ n/2 



Here LCM is the [least common multiple](https://en.wikipedia.org/wiki/Least_common_multiple) of numbers a_1, a_2, …, a_k.

We can show that for given constraints the answer always exists.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.

The only line of each test case contains two integers n, k (3 ≤ n ≤ 10^9, 3 ≤ k ≤ n).

It is guaranteed that the sum of k over all test cases does not exceed 10^5.

Output

For each test case print k positive integers a_1, a_2, …, a_k, for which all conditions are satisfied.

Example

Input


2
6 4
9 5


Output


1 2 2 1 
1 3 3 1 1 

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
#define F first
#define S second
#define EB emplace_back
#define MP make_pair
#define all(o) (o).begin(), (o).end()
#define mset(m,v) memset(m,v,sizeof(m))
#define rep(i,n) for(ll i=0;i<(n);++i)
#define pii pair<ll,ll>
#define repe(i,a,b) for(ll i=a;i<=b;++i)
#define revlp(i,a,b) for(ll i=a;i>=b;i--)
#define remin(a,b) (a=min((a),(b)))
#define remax(a,b) (a=max((a),(b))) 
#define sz(x) (ll)(x).size()
#define endl '\n'
typedef long long int ll;
typedef long double ld;
typedef pair<ll,ll> pll;     
typedef vector<ll> vi;
typedef vector<vi> graph;
long long mod=1000000007;     long double EPS=1e-9;
ll pow(ll x,ll y){ll res = 1;while(y){if(y&1)res = (res*x)%mod;y=y/2,x=(x*x)%mod;}return res%mod;}
#ifndef ONLINE_JUDGE
#define debarr(a,n)cerr<<#a<<":";for(int i=0;i<n;i++)cerr<<a[i]<<" ";cerr<<endl;
#define debmat(mat,row,col)cerr<<#mat<<":\n";for(int i=0;i<row;i++){for(int j=0;j<col;j++)cerr<<mat[i][j]<<" ";cerr<<endl;}
#define pr(...)dbs(#__VA_ARGS__,__VA_ARGS__)
template<class S,class T>ostream &operator<<(ostream &os,const pair<S,T> &p){return os<<"("<<p.first<<","<<p.second<<")";}
template<class T>ostream &operator<<(ostream &os,const vector<T> &p){os<<"[";for(auto&it:p)os<<it<<" ";return os<<"]";}
template<class T>ostream &operator<<(ostream &os,const set<T>&p){os<<"[";for(auto&it:p)os<<it<<" ";return os<<"]";}
template<class T>ostream &operator<<(ostream &os,const multiset<T>&p){os<<"[";for(auto&it:p)os<<it<<" ";return os<<"]";}
template<class S,class T>ostream &operator<<(ostream &os,const map<S,T>&p){os<<"[";for(auto&it:p)os<<it<<" ";return os<<"]";}
template<class T>void dbs(string str,T t){cerr<<str<<":"<<t<<"\n";}
template<class T,class...S>void dbs(string str,T t,S... s){int idx=str.find(',');cerr<<str.substr(0,idx)<<":"<<t<<",";dbs(str.substr(idx+1),s...);}
#else
#define pr(...){}
#define debarr(a,n){}
#define debmat(mat,row,col){},
#endif
#define io ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);


void solve(){
    int n,k;
    cin>>n>>k;
    if(k>3){
        rep(i,k-3){
            cout<<1<<" ";
            n--;
        }
    }

    if(n%2==1){
        cout<<n/2<<" "<<n/2<<" "<<1;
    }
    else{
        if(((n-2)/2)%2==0){
            cout<<(n-2)/2<<" "<<(n-2)/2<<" "<<2;       
        }
        else{
            cout<<n/2<<" "<<n/4<<" "<<n/4;
        }
    }
    cout<<endl;
}

    
int main(){    
    io;
    int test_Cases = 1;
    cin>>test_Cases;
    while(test_Cases--){    
      solve();
    }

}