Flip the Bits（思维）

You are given a positive integer n. Your task is to build a number m by flipping the minimum number of bits in the binary representation of n such that m is less than n (m < n) and it is as maximal as possible. Can you?

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 one integer n (1 ≤ n ≤ 10⁹), as described in the statement above.

Output

For each test case, print a single line containing the minimum number of bits you need to flip in the binary representation of n to build the number m.

Example

Input

2
5
10

Output

1
2


题目意思：将一个2进制的n中每个位翻转得到一个比n小且尽可能大的数，求输出翻转了几位。 

解题思路：这道题该由我背锅，我当时先是翻译错了题意，后来稍微有一点眉目了，我又理解错了那个flip的意思，这里面的翻转并不是那种交换(swap那样的），而是像硬币正面换到反面那样的翻转，也就
是0与1的交换，根据题意可以推出想要得到一个既要比n小还有尽可能大的数，只有是n前面的那一个数n-1。所以就是根据n构造一个二进制的n-1,方法就是找到n的二进制中最后面的那一个1翻转为0，而最
后一个1之后的0全都翻转成1，统计所用的翻转次数即可。

#include<cstdio>
#include<cstring>
int main()
{
    int t,n,j,k,i,count;
    int a[32];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        memset(a,-1,sizeof(a));
        j=0;
        count=0;
        i=n;
        while(i)
        {
            a[j]=i%2;
            if(a[j]==0)
            {
                count++;
            }
             if(a[j]==1)
             {
                 count++;
                 break;
             }
            i/=2;
            j++;
        }
        printf("%d
",count);
    }
    return 0;
}