Given an integer n, find the maximum value of integer k such that the following condition holds: 

n & (n-1) & (n-2) & (n-3) & ... (k) = 0  where & denotes the [bitwise AND operation.](https://en.wikipedia.org/wiki/Bitwise_operation#AND)

Input

The first line contains a single integer t (1 ≤ t ≤ 3 ⋅ 10^4). Then t test cases follow.

The first line of each test case contains a single integer n (1 ≤ n ≤ 10^9).

Output

For each test case, output a single integer — the required integer k.

Example

Input


3
2
5
17


Output


1
3
15

Note

In the first testcase, the maximum value for which the continuous & operation gives 0 value, is 1.

In the second testcase, the maximum value for which the continuous & operation gives 0 value, is 3. No value greater then 3, say for example 4, will give the & sum 0. 

  * 5   \&   4 ≠ 0, 
  * 5   \&   4   \&   3 = 0. 



Hence, 3 is the answer.
import java.io.*;
import java.util.*;

public class A {

	static int mod = 1000000007;

	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		int yo = sc.nextInt();
		while (yo-- > 0) {
			int n = sc.nextInt();
			if(n == 1) {
				out.println("0");
				continue;
			}
			String s = decToBinary(n);
//			System.out.println(s);
			int size = s.length()-1;
			StringBuilder sb = new StringBuilder();
			for(int i = 0; i < size; i++) {
				sb.append("1");
			}
			
			if(sb.length() == 0) {
				out.println("0");
			}
			else {
				out.println(Integer.parseInt(sb.toString(),2));
			}
			
			
		}
		out.close();
	}
	
	static String decToBinary(int n)
    {
        // array to store binary number
        int[] binaryNum = new int[1000];
   
        // counter for binary array
        int i = 0;
        while (n > 0) 
        {
            // storing remainder in binary array
            binaryNum[i] = n % 2;
            n = n / 2;
            i++;
        }
   
        String s = "";
        for (int j = i - 1; j >= 0; j--)
            s += (binaryNum[j])+"";
        
        return s;
    }

	static class Pair {
		int x;
		int y;

		public Pair(int x, int y) {
			this.x = x;
			this.y = y;
		}
	}

	static void ruffleSort(int[] a) {
		int n = a.length;
		Random r = new Random();
		for (int i = 0; i < a.length; i++) {
			int oi = r.nextInt(n), temp = a[i];
			a[i] = a[oi];
			a[oi] = temp;
		}
		Arrays.sort(a);
	}

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

	static boolean[] sieve(int n) {
		boolean isPrime[] = new boolean[n + 1];
		for (int i = 2; i <= n; i++) {
			if (isPrime[i])
				continue;
			for (int j = 2 * i; j <= n; j += i) {
				isPrime[j] = true;
			}
		}
		return isPrime;
	}

	static long pow(int a, long b) {
		if (b == 0) {
			return 1;
		}
		if (b == 1) {
			return a;
		}
		if (b % 2 == 0) {
			long ans = pow(a, b / 2);
			return ans * ans;
		} else {
			long ans = pow(a, (b - 1) / 2);
			return a * ans * ans;
		}

	}

	static class FastScanner {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");

		String next() {
			while (!st.hasMoreTokens())
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

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

		long nextLong() {
			return Long.parseLong(next());
		}
	}

	//	For Input.txt and Output.txt	
	//	FileInputStream in = new FileInputStream("input.txt");
	//	FileOutputStream out = new FileOutputStream("output.txt");
	//	PrintWriter pw = new PrintWriter(out);
	//	Scanner sc = new Scanner(in);
}