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 rep(i,a,b) for(i=a;i<b;i++)
#define pb push_back
typedef long long ll;
typedef long double ld;
const ll INF = 9223372036854775807;
const ll MOD = 1000000007;
const long double PI = acos(-1);
#define mp make_pair
#define all(v) v.begin(), v.end()
#define F first
#define S second
typedef vector<int> vi;
typedef map<int, int> mii;
typedef vector<vi> vvi;

 //debugger
void __print(int x) {cerr << x;}
void __print(long x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}

template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifndef ONLINE_JUDGE
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif

// debugger end
//global variables


//functions 
void solve()
{
    ll n,k;
    cin>>n>>k;
    if(k>3)
    {
        while(k!=3)
        {
            n--;
            k--;
            cout<<"1"<<" ";
        }
    }
    if(n%2==0)
    {   if(n%4!=0)
        cout<<(n-2)/2<<" "<<(n-2)/2<<" "<<"2"<<endl;
        else
            cout<<n/2<<" "<<n/4<<" "<<n/4<<endl;
    }else
    {
        cout<<(n-1)/2<<" "<<(n-1)/2<<" "<<"1"<<endl;
    }
}

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    int test;
   cin>>test;
   while(test--)
   {
       solve();
    }
 



 return 0;
}