Problem: You are given an array [a_1, a_2, ..., a_n] such that 1 ≤ a_i ≤ 10^9. Let S be the sum of all elements of the array a.

Let's call an array b of n integers beautiful if:

  * 1 ≤ b_i ≤ 10^9 for each i from 1 to n; 
  * for every pair of adjacent integers from the array (b_i, b_{i + 1}), either b_i divides b_{i + 1}, or b_{i + 1} divides b_i (or both); 
  * 2 ∑ _{i = 1}^{n} |a_i - b_i| ≤ S. 



Your task is to find any beautiful array. It can be shown that at least one beautiful array always exists.

Input

The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.

Each test case consists of two lines. The first line contains one integer n (2 ≤ n ≤ 50).

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).

Output

For each test case, print the beautiful array b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^9) on a separate line. It can be shown that at least one beautiful array exists under these circumstances. If there are multiple answers, print any of them.

Example

Input


4
5
1 2 3 4 5
2
4 6
2
1 1000000000
6
3 4 8 1 2 3


Output


3 3 3 3 3
3 6
1 1000000000
4 4 8 1 3 3

Solution (in JAVA):

Solution:
import java.util.*;
import java.lang.Math;
import java.util.concurrent.TimeUnit;
import java.util.stream.Collectors;
import java.io.*;
import java.math.BigInteger;
import java.io.FileWriter;

public class CodeForces1 {

	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());
		}

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

	static FastReader f = new FastReader();

	static void print(String s) {
		System.out.print(s);
	}

	static void print(boolean s) {
		System.out.print(s);
	}

	static boolean isPrime(long n) {

		if (n <= 1)
			return false;
		if (n <= 3)
			return true;

		if (n % 2 == 0 || n % 3 == 0)
			return false;

		for (long i = 5; i * i <= n; i = i + 6)
			if (n % i == 0 || n % (i + 2) == 0)
				return false;

		return true;
	}

	static long[] inArr(long a[], int n) {
		for (int i = 0; i < n; i++)
			a[i] = f.nextLong();
		return (a);
	}

	static void printSubsets(int set[]) {
		int n = set.length;
		Arrays.sort(set);
		for (int i = 0; i < n; i++) {

		}
		HashMap<Integer, Integer> hm = new HashMap<>();
		for (int i = 0; i < (1 << n); i++) {
			for (int j = 0; j < n; j++) {
				if ((i & (1 << j)) > 0) {
					if (hm.get(set[j]) == null) {
						hm.put(set[j], 0);
					}
					hm.put(set[j], hm.get(set[j]) + 1);

				}
			}

		}
	}

	static long gcd(long a, long b) {
		if (a == 0)
			return b;
		return gcd(b % a, a);
	}

	static long lcm(long a, long b) {
		return (a * b) / gcd(a, b);
	}

	static long power(long x, long y, long p) {

		// Initialize result
		long res = 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 % 2 == 1)
				res = (res * x) % p;

			// y must be even now
			y = y >> 1; // y = y/2
			x = (x * x) % p;
		}

		return res;
	}

	static long modInverse(long n, long p) {
		return power(n, p - 2, p);
	}

	static long nCrModPFermat(int n, int r) {
		long p = 1000000007;

		// Base case
		if (r == 0)
			return 1;

		// Fill factorial array so that we
		// can find all factorial of r, n
		// and n-r
		long[] fac = new long[n + 1];
		fac[0] = 1;

		for (int i = 1; i <= n; i++)
			fac[i] = fac[i - 1] * i % p;

		return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) % p;
	}
	List<Integer> removeDup(ArrayList<Integer> list){
		List<Integer> newList = list.stream() 
                                      .distinct() 
									  .collect(Collectors.toList()); 
		return newList;
	}
	static int lcm(int a, int b) {
		return (a / __gcd(a, b) * b);
	}

	static int __gcd(int a, int b) {
		return b == 0 ? a : __gcd(b, a % b);
	}

	// Function to find and print the two numbers
	static int[] findNums(int x) {

		int ans = -1;

		// To find the factors
		for (int i = 1; i <= Math.sqrt(x); i++) {

			// To check if i is a factor of x and
			// the minimum possible number
			// satisfying the given conditions
			if (x % i == 0 && lcm(i, x / i) == x) {

				ans = i;
			}
		}
		int a[] = { ans, x / ans };
		return (a);// (ans + " " + (x / ans));
	}

	static int solve(String a[], int row, int col1, int col2, /* int count, */int n, int m) {
		int flag = 0;
		for (int i = col1; i <= col2; i++) {
			if (a[row].charAt(i) == '*') {
				if (row + 1 < n && col1 - 1 >= 0 && col2 + 1 < m) {
					flag = solve(a, row + 1, col1 - 1, col2 + 1, n, m);
				} else {
					continue;
				}
			} else {
				flag = 0;
				return flag;
			}
		}
		flag++;
		return flag;

	}

	public static void main(String args[]) throws Exception {
		BufferedWriter w = new BufferedWriter(new OutputStreamWriter(System.out));
		Random rand = new Random();
		int t = f.nextInt();
		while (t-- != 0) {
			int n=f.nextInt();
			//int p=50000000000;
			long sum=0;
			int a[]=new int[n];
			for(int i=0;i<n;i++){
				a[i]=f.nextInt();
				sum+=a[i];
			}
			int b1[]=new int[n];
			int b2[]=new int[n];
			long sum1=0,sum2=0;
			for(int i=0;i<n;i++){
				b1[i]=i%2==0?1:a[i];
				sum1+=Math.abs(b1[i]-a[i]);
				b2[i]=i%2==0?a[i]:1;
				sum2+=Math.abs(b2[i]-a[i]);
			}
			if(2*sum1<=sum){
				for(int i=0;i<n;i++){
					w.write(b1[i]+" ");
				}
			}
			else if(2*sum2<=sum){
				for(int i=0;i<n;i++){
					w.write(b2[i]+" ");
				}
			}
			w.write("\n");	

		}
		w.flush();

	}
}