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.
import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.StringTokenizer;

/**
 * Created by timur on 28.03.15.
 */

public class TaskC {
    boolean eof;
    BufferedReader br;
    StringTokenizer st;
    PrintWriter out;

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

    public String nextToken() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                eof = true;
                return "-1";
            }
        }
        return st.nextToken();
    }

    public int nextInt() {
        return Integer.parseInt(nextToken());
    }

    public long nextLong() {
        return Long.parseLong(nextToken());
    }

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

    String nextLine() throws IOException {
        return br.readLine();
    }


    void run() throws IOException {
        InputStream input = System.in;
        PrintStream output = System.out;
        try {
            File f = new File("a.in");
            if (f.exists() && f.canRead()) {
                input = new FileInputStream(f);
                output = new PrintStream("a.out");
            }
        } catch (Throwable e) {
        }
        br = new BufferedReader(new InputStreamReader(input));
        out = new PrintWriter(output);
        solve();
        br.close();
        out.close();
    }

    class Pair implements Comparable<Pair> {
        int l, r, ind;

        public Pair(int l, int r, int ind) {
            this.l = l;
            this.r = r;
            this.ind = ind;
        }

        @Override
        public int compareTo(Pair o) {
            int r = l - o.l;
            if (r == 0)
                r = o.r - this.r;
            return r;
        }
    }

    int gcd(int a, int b) {
        if (b == 0)
            return a;
        else
            return gcd(b, a % b);
    }
    void solve() {
        int n = nextInt();
        int[] p = new int[30];
        p[0] = 1;
        for (int i = 1; i < 30; i++) {
            p[i] = p[i - 1] * 2 + 1;
//            out.println(p[i]);
        }
        int[] ans = {0, 1, 1, 5, 1,
            21,
            1,
            85,
            73,
            341,
            89,
            1365,
            1,
            5461,
            4681,
            21845,
            1,
            87381,
            1,
            349525,
            299593,
            1398101,
            178481,
            5592405,
            1082401,
            22369621,
        1};
        int x, res, a, b;
        for (int i = 0; i < n; i++) {
            x = nextInt();

            for (int j = 0; j < 30; j++) {
                if (x == p[j]) {

//                    if (j % 2 == 0)
//                        out.println("1");
//                    else
//                        out.println(x / 3);
//                    res = 0;
//                    for (a = 1; a < x; a++) {
//                        res = Math.max(res, gcd(x ^ a, x & a));
//                    }
                    out.println(ans[j]);
                    break;
                } else if (x < p[j]) {
                    out.println(p[j]);
                    break;
                }
            }
        }
    }
}