Yousef loves playing with functions in his free time. Today, he invents the following function:

Yousef will give you a list of queries, and you need to find the answers for them. For each query, you are given an integer n, and your task is to count the number of integers m in which (0 ≤ m

Input

The first line contains an integer T (1 ≤ T ≤ 105) specifying the number of test cases.

Each test case consists of a single line containing an integer n (0 ≤ n ≤ 1018), as described in the problem statement above.

Output

For each test case, print a single line containing the number of integers m in which (0 ≤ m ≤ n) and calc(n,  m) is an even number.

Example

Input

2
1
2

Output

0
1

題意：C[i][j]  j<=i  的偶數有多少個

題解：打表偶數的找不到規律，那就看看奇數的啊，發現和2進製為1的個數有關

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int N=1e5+100;
typedef long long ll;
const ll mod=1e9+7;
ll C[1100][1100];
int main()
{
	/*
	C[0][0]=C[1][0]=C[1][1]=1;
	for(int i=2;i<=1000;i++)
	{
		C[i][0]=1;
		int cnt=1;
		for(int j=1;j<=i;j++)
		{
			C[i][j]=(C[i-1][j]+C[i-1][j-1])%1000000;
			if(C[i][j]&1) cnt++;	
		}
		printf("%d: %d\n",i,cnt);
	}
	*/
	int T;
	scanf("%d",&T);
	while(T--)
	{
		ll n,m,cnt=0;
		scanf("%lld",&n);
		m=n;
		while(m)
		{
			cnt++;
			m-=m&(-m);
		}
		printf("%lld\n",n+1-(1ll<<cnt));
	}
	return 0;
}