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:
#pragma GCC optimize "trapv"
#include<iostream>
#include <bits/stdc++.h>
using namespace std;
#define fio ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define ll long long
#define ull unsigned long long
#define ui unsigned int
#define vi vector<int>
#define vll vector<ll>
#define pb push_back
#define ld long double
#define mp make_pair
#define pii pair<int,int>
#define mod 1000000007
#define rep(i,n) for(int i=0;i<n;i++)
#define repp(i,a,n) for(int i=a;i<n;i++)
#define all(v) v.begin(),v.end()
#define input(arr,n) for(ll i1=0;i1<n;i1++ )cin>>arr[i1]
vector<string> split(const string& s, char c) {
    vector<string> v; stringstream ss(s); string x;
    while (getline(ss, x, c)) v.push_back(x); return move(v);
}
template<typename T, typename... Args>
inline string arrStr(T arr, int n) {
    stringstream s; s << "[";
    for(int i = 0; i < n - 1; i++) s << arr[i] << ",";
    s << arr[n - 1] << "]";
    return s.str();
}
#define EVARS(args...) {__evars_begin(__LINE__); __evars(split(#args, ',').begin(), args);}
inline void __evars_begin(int line) { cerr << "#" << line << ": "; }
template<typename T> inline void __evars_out_var(vector<T> val) { cerr << arrStr(val, val.size()); }
template<typename T> inline void __evars_out_var(T* val) { cerr << arrStr(val, 10); }
template<typename T> inline void __evars_out_var(T val) { cerr << val; }
inline void __evars(vector<string>::iterator it) { cerr << endl; }
template<typename T, typename... Args>
inline void __evars(vector<string>::iterator it, T a, Args... args) {
    cerr << it->substr((*it)[0] == ' ', it->length()) << "=";
    __evars_out_var(a);
    cerr << "; ";
    __evars(++it, args...);
}
template<typename T_vector>
void output_vector(const T_vector &v, bool add_one = false, int start = -1, int end = -1) {
    if (start < 0) start = 0;
    if (end < 0) end = int(v.size());
 
    for (int i = start; i < end; i++)
        cout << v[i] + (add_one ? 1 : 0) << (i < end - 1 ? ' ' : '\n');
}
struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
        // http://xorshift.di.unimi.it/splitmix64.c
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }

    size_t operator()(uint64_t x) const {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};
template<class T, class H>using umap=unordered_map<T,H,custom_hash>;
template<class T>using uset=unordered_set<T,custom_hash>;
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif
    fio;
    clock_t clk = clock();
    int t = 1;
    cin >> t;
    while(t--)
    {
        ll n, k;
        cin >> n >> k;
        if(k > 3)
        {
            k -= 3;
            for(int i = 0; i < k; i++)
            cout << 1 << " ";
            n -= k;
        }
        if(n % 2 != 0)
        {
            cout << n / 2 << " " << n / 2 << " " << 1 << endl;
        }
        else if(n % 4 == 0)
        {
            cout << n / 2 << " " << n / 4 << " " << n / 4 << endl;
        }
        else
        {
            cout << n / 2 - 1 << " " << n / 2 - 1 << " " << 2 << endl;
        }
    }
    cerr << '\n'<<"Time (in s): " << double(clock() - clk) * 1.0 / CLOCKS_PER_SEC << '\n';    
}