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 <iostream>
#include <iomanip>
#include <string>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
#include <numeric>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#include <random>
#include <queue>
#include <memory>
#include <fstream>
#include <iterator>
#include <tuple>
#include <random>
#include <functional>
#include <cassert>
#include <chrono> 
#include <array>
#include <stack>
 
using namespace std;

template <typename A, typename B>
string to_string(pair<A, B> p);
 
template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p);
 
template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p);
 
string to_string(const string& s) {
  return '"' + s + '"';
}
 
string to_string(const char* s) {
  return to_string((string) s);
}
 
string to_string(bool b) {
  return (b ? "true" : "false");
}
 
string to_string(vector<bool> v) {
  bool first = true;
  string res = "{";
  for (int i = 0; i < static_cast<int>(v.size()); i++) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(v[i]);
  }
  res += "}";
  return res;
}
 
template <size_t N>
string to_string(bitset<N> v) {
  string res = "";
  for (size_t i = 0; i < N; i++) {
    res += static_cast<char>('0' + v[i]);
  }
  return res;
}
 
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto &x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
 
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
 
template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " + to_string(get<2>(p)) + ")";
}
 
template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " + to_string(get<2>(p)) + ", " + to_string(get<3>(p)) + ")";
}
 
void debug_out() { cerr << endl; }
 
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
 
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endif

typedef long double ld;
typedef long long ll;
 
#define x first
#define y second
#define pb push_back
#define int long long
#define eb emplace_back
#define sz(x) (int)x.size()
#define all(x) x.begin(), x.end()
 
const ll INF = 1e18 + 7;
const int N = 3e5 + 5;
const ld EPS = 1e-12;
const int MOD = 1e9 + 7;

void solve() {
    int n, k;
    cin >> n >> k;
    for (int i = 0; i < k - 3; i++) {
        cout << 1 << ' ';
    }
    n -= k - 3;
    if (n % 4 == 0) {
        cout << n / 2 << ' ' << n / 4 << ' ' << n / 4 << '\n';
    } else if (n % 4 == 2) {
        cout << 2 << ' ' << (n - 2) / 2 << ' ' << (n - 2) / 2 << '\n';
    } else {
        cout << 1 << ' ' << n / 2 << ' ' << n / 2 << '\n';
    }
}


signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
 
    int tests = 1;
    cin >> tests;
 
    while (tests-- > 0) {
        solve();
    }
 
    return 0;
}