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 JAVA):

Solution:
import java.io.*;
import java.util.*;
import java.math.*;

import static java.awt.geom.Point2D.distance;
                    /*       :)  :)  :)  :)                */
public class Main {

    static FastReader in;
    static PrintWriter o;

    public static void solve() {
        int t = in.nextInt();
        while (t-- > 0) {
            int n = in.nextInt();
            int k = in.nextInt();
            int n1 = n - 3;
            int k1 = k - 3;
            if (k == 3) {
                int[] x = find(n, k);
                o.println(x[0] + " " + x[1] + " " + x[2]);
                continue;
            }
            for (int i = 0; i < k1; i++) {
                o.print("1 ");
            }
            if (n1 != 0 && k1 != 0) {
                int[] x = find(n-(k-3), 3);
                o.print(x[0] + " " + x[1] + " " + x[2]);
            }
            o.println();
        }
        o.close();
        return;
    }

    public static int[] find(int n, int k) {
        int x1 = -1;
        int x2 = -1;
        int x3 = -1;
        if (n % 2 == 0) {
            int v = n / 2;
            if (v % 2 == 1) {
                x1 = 2;
                x2 = (n-2)/2;
                x3 = (n-2)/2;
            } else {
                int x = n / 4;
                x1 = 2 * x;
                x2 = x;
                x3 = x;
            }
        } else {
            x1 = 1;
            x2 = n/2;
            x3 = n/2;
        }
        return new int[] {x1, x2, x3};
    }

    public static void main(String[] args) {
        in = new FastReader();
        o = new PrintWriter(System.out);
        solve();
        return;
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }

        long[] readLongArray(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++) a[i] = nextLong();
            return a;
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

        boolean isReady() throws Throwable{
            return br.ready();
        }

    }

}