Problem: You have n sticks of the given lengths.

Your task is to choose exactly four of them in such a way that they can form a rectangle. No sticks can be cut to pieces, each side of the rectangle must be formed by a single stick. No stick can be chosen multiple times. It is guaranteed that it is always possible to choose such sticks.

Let S be the area of the rectangle and P be the perimeter of the rectangle. 

The chosen rectangle should have the value (P^2)/(S) minimal possible. The value is taken without any rounding.

If there are multiple answers, print any of them.

Each testcase contains several lists of sticks, for each of them you are required to solve the problem separately.

Input

The first line contains a single integer T (T ≥ 1) — the number of lists of sticks in the testcase.

Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th list. The first line of the pair contains a single integer n (4 ≤ n ≤ 10^6) — the number of sticks in the i-th list. The second line of the pair contains n integers a_1, a_2, ..., a_n (1 ≤ a_j ≤ 10^4) — lengths of the sticks in the i-th list.

It is guaranteed that for each list there exists a way to choose four sticks so that they form a rectangle.

The total number of sticks in all T lists doesn't exceed 10^6 in each testcase.

Output

Print T lines. The i-th line should contain the answer to the i-th list of the input. That is the lengths of the four sticks you choose from the i-th list, so that they form a rectangle and the value (P^2)/(S) of this rectangle is minimal possible. You can print these four lengths in arbitrary order.

If there are multiple answers, print any of them.

Example

Input

3
4
7 2 2 7
8
2 8 1 4 8 2 1 5
5
5 5 5 5 5


Output

2 7 7 2
2 2 1 1
5 5 5 5

Note

There is only one way to choose four sticks in the first list, they form a rectangle with sides 2 and 7, its area is 2 ⋅ 7 = 14, perimeter is 2(2 + 7) = 18. (18^2)/(14) ≈ 23.143.

The second list contains subsets of four sticks that can form rectangles with sides (1, 2), (2, 8) and (1, 8). Their values are (6^2)/(2) = 18, (20^2)/(16) = 25 and (18^2)/(8) = 40.5, respectively. The minimal one of them is the rectangle (1, 2).

You can choose any four of the 5 given sticks from the third list, they will form a square with side 5, which is still a rectangle with sides (5, 5).

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
import java.lang.*;
public class TwoTv {
	public static void main(String[] args) throws FileNotFoundException {
		try
		{
			InputReader in = new InputReader(System.in);
			PrintWriter out = new PrintWriter(System.out);
			int t=in.nextInt();
			for(int i=0;i<t;i++){
				int n=in.nextInt();
				int a[]=new int[n];
				for(int j=0;j<n;j++)
					a[j]=in.nextInt();
				//for(int j=0;j<n;j++)
				//	System.out.println(a[j]+" ");
				sort(a,0,n-1);
				//for(int j=0;j<n;j++)
				//	System.out.println(a[j]+" ");
				int f[]=new int[n/2];
				int k=0;
				for(int j=0;j<n-1;j++){
					//System.out.print(a[j]+" ");
					if(a[j]==a[j+1]){
						f[k]=a[j];
						j++;
						k++;
					//	System.out.println("ch "+f[k-1]+" ch");
					}
					//System.out.println();
					if(k>1&&(f[k-1]==f[k-2])){
						out.println(f[k-1]+" "+f[k-1]+" "+f[k-2]+" "+f[k-2]);
						break;
					}
				}
				//System.out.println();
				if(k>1&&(f[k-1]==f[k-2])){
					continue;
				}
				int m1=0,m2=0;
				double min=1000000000;
				for(int j=k-1;j>0;j--){
					//System.out.println(f[j]+" hello "+f[j-1]);
					double temp=((double) f[j])/(f[j-1]+0.0);
					if(temp<min) {
						m1=f[j];
						m2=f[j-1];
						min=temp;
						//System.out.println(f[j]+" w8 for it "+f[j-1]+" "+min);
					}
				}
				out.println(m1+" "+m1+" "+m2+" "+m2);
			}
			out.close();
		}
		catch(Exception e){
		}
	}
	public static int comb(int a,int b){
		long prod=1;
		for(int i=a;i>Math.max(b,a-b);i--){
			prod*=i;
		}
		for(int i=2;i<=Math.min(b,a-b);i++){
			prod/=i;
		}
		return (int) (prod%1000000007);
	}
	public static long gcd(long a, long b)
	{
		if (a == 0)
			return b;

		return gcd(b%a, a);
	}
	public static int search(int arr[], int l, int h, int key)
	{
		if (l > h)
			return -1;
		int mid = (l+h)/2;
		if (arr[mid] == key)
			return mid;
		if (arr[l] <= arr[mid])
		{
			if (key >= arr[l] && key <= arr[mid])
				return search(arr, l, mid-1, key);

			return search(arr, mid+1, h, key);
		}
		if (key >= arr[mid] && key <= arr[h])
			return search(arr, mid+1, h, key);

		return search(arr, l, mid-1, key);
	}
	public static long power(long x, long y, long p)
	{
		// Initialize result
		long res = (long)1;

		// Update x if it is more
		// than or equal to p
		x = x % p;

		while (y > 0)
		{
			// If y is odd, multiply x
			// with result
			if((y & 1)==1)
				res = (res * x) % p;

			// y must be even now
			// y = y / 2
			y = y >> 1;
			x = (x * x) % p;
		}
		return res;
	}
	public static void merge(int arr[], int l, int m, int r)
	{
		int n1 = m - l + 1;
		int n2 = r - m;
		int L[] = new int [n1];
		int R[] = new int [n2];
		for (int i=0; i<n1; ++i)
			L[i] = arr[l + i];
		for (int j=0; j<n2; ++j)
			R[j] = arr[m + 1+ j];
		int i = 0, j = 0;
		int k = l;
		while (i < n1 && j < n2)
		{
			if (L[i] <= R[j])
			{
				arr[k] = L[i];
				i++;
			}
			else
			{
				arr[k] = R[j];
				j++;
			}
			k++;
		}
		while (i < n1)
		{
			arr[k] = L[i];
			i++;
			k++;
		}
		while (j < n2)
		{
			arr[k] = R[j];
			j++;
			k++;
		}
	}
	public static void sort(int arr[], int l, int r)
	{
		if (l < r)
		{
			int m = (l+r)/2;
			sort(arr, l, m);
			sort(arr , m+1, r);
			merge(arr, l, m, r);
		}
	}
	static class InputReader {

		private final InputStream stream;
		private final byte[] buf = new byte[8192];
		private int curChar, snumChars;

		public InputReader(InputStream stream) {
			this.stream = stream;
		}

		public int read() {
			if (snumChars == -1)
				throw new InputMismatchException();
			if (curChar >= snumChars) {
				curChar = 0;
				try {
					snumChars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (snumChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}

		public int nextInt() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			int res = 0;
			do {
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}

		public long nextLong() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			long res = 0;
			do {
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}

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

		public String readString() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}

		public String nextLine() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isEndOfLine(c));
			return res.toString();
		}

		public boolean isSpaceChar(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

		private boolean isEndOfLine(int c) {
			return c == '\n' || c == '\r' || c == -1;
		}

	}
}