Can the greatest common divisor and bitwise operations have anything in common? It is time to answer this question.

Suppose you are given a positive integer a. You want to choose some integer b from 1 to a - 1 inclusive in such a way that the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers a ⊕ b and a \> \& \> b is as large as possible. In other words, you'd like to compute the following function:

$$$f(a) = max_{0 < b < a}{gcd(a ⊕ b, a \> \& \> b)}.$$$

Here ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR), and \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).

The greatest common divisor of two integers x and y is the largest integer g such that both x and y are divided by g without remainder.

You are given q integers a_1, a_2, …, a_q. For each of these integers compute the largest possible value of the greatest common divisor (when b is chosen optimally). 

Input

The first line contains an integer q (1 ≤ q ≤ 10^3) — the number of integers you need to compute the answer for.

After that q integers are given, one per line: a_1, a_2, …, a_q (2 ≤ a_i ≤ 2^{25} - 1) — the integers you need to compute the answer for. 

Output

For each integer, print the answer in the same order as the integers are given in input.

Example

Input


3
2
3
5


Output


3
1
7

Note

For the first integer the optimal choice is b = 1, then a ⊕ b = 3, a \> \& \> b = 0, and the greatest common divisor of 3 and 0 is 3.

For the second integer one optimal choice is b = 2, then a ⊕ b = 1, a \> \& \> b = 2, and the greatest common divisor of 1 and 2 is 1.

For the third integer the optimal choice is b = 2, then a ⊕ b = 7, a \> \& \> b = 0, and the greatest common divisor of 7 and 0 is 7.
//package a2oj_ladder_C;

import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.InputMismatchException;
import java.util.LinkedList;
import java.util.Queue;
import java.util.TreeMap;
import java.util.TreeSet;


@SuppressWarnings("unused")
public class GlobalRound1C {

	InputStream is;
	PrintWriter out;
	String INPUT = "";

	void swap(int a[] , int l , int i) {
		int t = a[i];
		a[i] = a[l];
		a[l] = t;		
	}
	void heapify(int a[] , int n , int i) {

		int l = 2*i+1;
		int r = 2*i+2;
		int largest = i;

		if(l<n && a[l]>a[i]) largest = l;
		if(r<n && a[r]>a[largest]) largest = r;

		if(largest != i) {
			swap(a , largest , i);
			heapify(a,n,largest);
		}		

	}

	int mod = (int)(Math.pow(10,9)+7);

	static boolean isPrime(int n) 
    { 
        // Corner cases 
        if (n <= 1) 
            return false; 
        if (n <= 3) 
            return true; 
  
        // This is checked so that we can skip 
        // middle five numbers in below loop 
        if (n % 2 == 0 || n % 3 == 0) 
            return false; 
  
        for (int i = 3; i * i <= n; i++) 
            if (n % i == 0) 
                return false; 
  
        return true; 
    } 
	
	void solve()
	{	
		int q = ni();
		int a[] = na(q);
		
		
		for(int i : a) {
			int j = 0;
			while((1<<j) <= i) {
				j++;
			}
			
			if(i == (1<<j)-1) {
				if(isPrime(i)) out.println(1);
				else {
					for(int k = 2 ; k*k<=i ; k++) {
						if(i%k == 0) {
							out.println(i/k);
							break;
						}
					}
				}
			}
			else {
				out.println((1<<j)-1);
			}
		}

	}


	void run() throws Exception
	{
		is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
		out = new PrintWriter(System.out);

		long s = System.currentTimeMillis();
		solve();
		out.flush();
		if(!INPUT.isEmpty())tr(System.currentTimeMillis()-s+"ms");
	}

	public static void main(String[] args) throws Exception { new GlobalRound1C().run(); }

	private byte[] inbuf = new byte[1024];
	public int lenbuf = 0, ptrbuf = 0;

	private int readByte()
	{
		if(lenbuf == -1)throw new InputMismatchException();
		if(ptrbuf >= lenbuf){
			ptrbuf = 0;
			try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
			if(lenbuf <= 0)return -1;
		}
		return inbuf[ptrbuf++];
	}

	private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
	private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }

	private double nd() { return Double.parseDouble(ns()); }
	private char nc() { return (char)skip(); }

	private String ns()
	{
		int b = skip();
		StringBuilder sb = new StringBuilder();
		while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ')
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}

	private char[] ns(int n)
	{
		char[] buf = new char[n];
		int b = skip(), p = 0;
		while(p < n && !(isSpaceChar(b))){
			buf[p++] = (char)b;
			b = readByte();
		}
		return n == p ? buf : Arrays.copyOf(buf, p);
	}

	private char[][] nm(int n, int m)
	{
		char[][] map = new char[n][];
		for(int i = 0;i < n;i++)map[i] = ns(m);
		return map;
	}

	private int[] na(int n)
	{
		int[] a = new int[n];
		for(int i = 0;i < n;i++)a[i] = ni();
		return a;
	}

	private int[][] na(int n , int m)
	{
		int[][] a = new int[n][m];
		for(int i = 0;i < n;i++) 
			for(int j = 0 ; j<m ; j++) a[i][j] = ni();
		return a;
	}

	private int ni()
	{
		int num = 0, b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}

		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}

	private long nl()
	{
		long num = 0;
		int b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}

		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}

	void display2D(int a[][]) {
		for(int i[] : a) {
			for(int j : i) {
				out.print(j+" ");
			}
			out.println();
		}
	}

	int[][] creategraph(int n , int m) {
		int g[][] = new int[n+1][];

		int from[] = new int[m];
		int to[] = new int[m];
		int ct[] = new int[n+1];

		for(int i = 0 ; i<m; i++) {
			from[i] = ni();
			to[i] = ni();
			ct[from[i]]++;
			ct[to[i]]++;
		}

		int parent[] = new int[n+1];

		for(int i = 0 ; i<n+1 ; i++) g[i] = new int[ct[i]];

		for(int i = 0 ; i<m ; i++) {
			g[from[i]][--ct[from[i]]] = to[i];
			g[to[i]][--ct[to[i]]] = from[i];
		}

		return g;
	}

	static long __gcd(long a, long b)
	{

		if(b == 0) 
		{
			return a;
		}
		else
		{
			return __gcd(b, a % b);
		}
	}

	// To compute x^y under modulo m
	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;
	}

	// Function to find modular 
	// inverse of a under modulo m
	// Assumption: m is prime
	static long modInverse(long a, int m)
	{
		if (__gcd(a, m) != 1) {
			//System.out.print("Inverse doesn't exist");
			return -1;
		}

		else {

			// If a and m are relatively prime, then
			// modulo inverse is a^(m-2) mode m
			//			System.out.println("Modular multiplicative inverse is "
			//											+power(a, m - 2, m));
			return power(a, m - 2, m);
		}
	}

	static long nCrModPFermat(int n, int r,
			int p , long fac[])
	{

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

		// Fill factorial array so that we
		// can find all factorial of r, n
		// and n-r
		long t = (fac[n]* modInverse(fac[r], p))%p;

		return ( (t* modInverse(fac[n-r], p))% p);
	}


	private static void tr(Object... o) { System.out.println(Arrays.deepToString(o)); }
}
