Tomorrow is a difficult day for Polycarp: he has to attend a lectures and b practical classes at the university! Since Polycarp is a diligent student, he is going to attend all of them.

While preparing for the university, Polycarp wonders whether he can take enough writing implements to write all of the lectures and draw everything he has to during all of the practical classes. Polycarp writes lectures using a pen (he can't use a pencil to write lectures!); he can write down c lectures using one pen, and after that it runs out of ink. During practical classes Polycarp draws blueprints with a pencil (he can't use a pen to draw blueprints!); one pencil is enough to draw all blueprints during d practical classes, after which it is unusable.

Polycarp's pencilcase can hold no more than k writing implements, so if Polycarp wants to take x pens and y pencils, they will fit in the pencilcase if and only if x + y ≤ k.

Now Polycarp wants to know how many pens and pencils should he take. Help him to determine it, or tell that his pencilcase doesn't have enough room for all the implements he needs tomorrow!

Note that you don't have to minimize the number of writing implements (though their total number must not exceed k).

Input

The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then the test cases follow.

Each test case is described by one line containing five integers a, b, c, d and k, separated by spaces (1 ≤ a, b, c, d, k ≤ 100) — the number of lectures Polycarp has to attend, the number of practical classes Polycarp has to attend, the number of lectures which can be written down using one pen, the number of practical classes for which one pencil is enough, and the number of writing implements that can fit into Polycarp's pencilcase, respectively.

In hacks it is allowed to use only one test case in the input, so t = 1 should be satisfied.

Output

For each test case, print the answer as follows:

If the pencilcase can't hold enough writing implements to use them during all lectures and practical classes, print one integer -1. Otherwise, print two non-negative integers x and y — the number of pens and pencils Polycarp should put in his pencilcase. If there are multiple answers, print any of them. Note that you don't have to minimize the number of writing implements (though their total number must not exceed k).

Example

Input


3
7 5 4 5 8
7 5 4 5 2
20 53 45 26 4


Output


7 1
-1
1 3

Note

There are many different answers for the first test case; x = 7, y = 1 is only one of them. For example, x = 3, y = 1 is also correct.

x = 1, y = 3 is the only correct answer for the third test case.
import java.util.*;
import java.io.*;

public class Main {
	static long mod = 1000000007;
	static int INF = 1000000000;
 	public static void main(String[] args){
		FastScanner scanner = new FastScanner();
		int t = scanner.nextInt();
		for(int query = 0; query < t; query++){
			int a = scanner.nextInt();
			int b = scanner.nextInt();
			int c = scanner.nextInt();
			int d = scanner.nextInt();
			int k = scanner.nextInt();
			int ans1 = -1;
			int ans2 = -1;
			for(int i = 1; i < 300; i++){
				for(int j = 1; j < 300; j++){
					if(i+j > k) continue;
					if(i*c >= a && j*d >= b){
						ans1 = i; ans2 = j;
						break;
					}
				}
			}
			if(ans1 == -1 && ans2 == -1){
				System.out.println(-1);
			}else{
				System.out.println(ans1+ " " + ans2);
			}
		}
	}
	static class BIT{
		int n;
		int[] bit;
		public BIT(int n){
			this.n = n;
			bit = new int[n+1];
		}
		void add(int idx, int val){
			for(int i = idx+1; i <= n; i += i&(-i)) bit[i-1] += val;
		}
		int sum(int idx){
			int res = 0;
			for(int i = idx+1; i > 0; i -= i&(-i)) res += bit[i-1];
			return res;
		}
		int sum(int begin, int end){
			if(begin == 0) return sum(end);
			return sum(end)-sum(begin-1);
		}
	}
	static class Pair implements Comparable<Pair>{
    int first, second;
    Pair(int a, int b){
        first = a;
        second = b;
    }
    @Override
    public boolean equals(Object o){
        if (this == o) return true;
        if (!(o instanceof Pair)) return false;
        Pair p = (Pair) o;
        return first == p.first && second == p.second;
    }
    @Override
    public int compareTo(Pair p){
        //return first == p.first ? second - p.second : first - p.first; //firstで昇順にソート
        //return (first == p.first ? second - p.second : first - p.first) * -1; //firstで降順にソート
        //return second == p.second ? first - p.first : second - p.second;//secondで昇順にソート
        //return (second == p.second ? first - p.first : second - p.second)*-1;//secondで降順にソート
		    //return first * 1.0 / second > p.first * 1.0 / p.second ? 1 : -1; // first/secondの昇順にソート
		    //return first * 1.0 / second < p.first * 1.0 / p.second ? 1 : -1; // first/secondの降順にソート
				//return second * 1.0 / first > p.second * 1.0 / p.first ? 1 : -1; // second/firstの昇順にソート
		    //return second * 1.0 / first < p.second * 1.0 / p.first ? 1 : -1; // second/firstの降順にソート
				//return Math.atan2(second, first) > Math.atan2(p.second, p.first) ? 1 : -1; // second/firstの昇順にソート
				//return first + second > p.first + p.second ? 1 : -1; //first+secondの昇順にソート
				//return first + second < p.first + p.second ? 1 : -1; //first+secondの降順にソート
				//return first - second < p.first - p.second ? 1 : -1; //first-secondの降順にソート
				return Math.atan2(second,first) > Math.atan2(p.second, p.first) ? 1 : -1;
		}
  }

	private static class FastScanner {
		private final InputStream in = System.in;
		private final byte[] buffer = new byte[1024];
		private int ptr = 0;
		private int buflen = 0;
		private boolean hasNextByte() {
				if (ptr < buflen) {
						return true;
				}else{
						ptr = 0;
						try {
								buflen = in.read(buffer);
						} catch (IOException e) {
								e.printStackTrace();
						}
						if (buflen <= 0) {
								return false;
						}
				}
				return true;
		}
		private int readByte() { if (hasNextByte()) return buffer[ptr++]; else return -1;}
		private static boolean isPrintableChar(int c) { return 33 <= c && c <= 126;}
		public boolean hasNext() { while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++; return hasNextByte();}
		public String next() {
				if (!hasNext()) throw new NoSuchElementException();
				StringBuilder sb = new StringBuilder();
				int b = readByte();
				while(isPrintableChar(b)) {
						sb.appendCodePoint(b);
						b = readByte();
				}
				return sb.toString();
		}
		public long nextLong() {
				if (!hasNext()) throw new NoSuchElementException();
				long n = 0;
				boolean minus = false;
				int b = readByte();
				if (b == '-') {
						minus = true;
						b = readByte();
				}
				if (b < '0' || '9' < b) {
						throw new NumberFormatException();
				}
				while(true){
						if ('0' <= b && b <= '9') {
								n *= 10;
								n += b - '0';
						}else if(b == -1 || !isPrintableChar(b)){
								return minus ? -n : n;
						}else{
								throw new NumberFormatException();
						}
						b = readByte();
				}
		}
		public int nextInt() {
				long nl = nextLong();
				if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) throw new NumberFormatException();
				return (int) nl;
		}
		public double nextDouble() { return Double.parseDouble(next());}
	}
}
