不大于 $n$ 且 $\operatorname{lowbit}$ 值不小于 $2^t$ 的数字个数为：

对 $\operatorname{lowbit}$ 值差分后累加即可。

#include <bits/stdc++.h>
using namespace std;
inline __int128 read()
{
	__int128 x = 0;
	char c = getchar_unlocked();
	while (c < '0') c = getchar_unlocked();
	while (c >= '0')
	{
		x = x * 10 + c - '0';
		c = getchar_unlocked();
	}
	return x;
}
void write(__int128 x)
{
	if (x <= 9)
	{
		putchar_unlocked(x + '0');
		return;
	}
	write(x / 10);
	putchar_unlocked(x % 10 + '0');
}
inline void writeline(__int128 x)
{
	write(x);
	putchar_unlocked('\n');
}
int main()
{
	int T = read();
	while (T--)
	{
		long long n = read();
		__int128 a = n;
		for (long long i = 1; (n >> i) >= 1; i++)
			a += n >> i << (i - 1);
		writeline(a);
	}
	return 0;
}